Bank Soal Latihan Limit Fungsi Aljabar

Sifat-Sifat Limit Fungsi

apabila k suatu konstanta, f dan g merupakan fungsi-fungsi yang mempunyai limit untuk x → a, a ∈ R maka berlaku:

Sifat-Sifat Limit Fungsi:

● $\lim_{ x\to 0}$ k = k

● $\lim_{ x\to 0}$ f(x) = f(a)

● $\lim_{ x\to 0}$ k.f(x) = k.$\lim_{ x\to 0}$ f(x)

● $\lim_{ x\to 0}$ {f(x) ± g(x)} = $\lim_{ x\to 0}$ f(x) ± $\lim_{ x\to 0}$ g(x)

● $\lim_{ x\to 0}$ {f(x) ± g(x)} = $\lim_{ x\to 0}$ f(x) . $\lim_{ x\to 0}$ g(x)

● $\lim_{ x\to 0}$ $\frac {f(x)}{g(x)}$ = $\frac {\lim_{ x\to 0} f(x)}{\lim_{ x\to 0} g(x)}$, untuk $\lim_{ x\to 0}$ ≠ 0

● $\lim_{ x\to 0}$ ${(f(x))^n}$ = $(\lim_{ x\to 0} f(x))^n$

Pembahasan Soal Limit Fungsi Aljabar

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1. $\lim_{ x\to a}$$\frac {x^2 + (3 - a)x - 3a}{x - a}$ = .....
a. a
b. a + 1
c. a + 2
d. a + 3
e. a + 4

penyelesaian:
$\lim_{ x\to a}$$\frac {x^2 + (3 - a)x - 3a}{x - a}$ = $\lim_{ x\to a}$$\frac {(x-a)(x-3)}{x + a}$
↔ = $\lim_{ x\to a}$ x + 3
↔ = a + 3

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2. $\lim_{ x\to 2}$$\frac {x^3 - 4x}{x - 2}$ = .....
a. 2
b. 4
c. 6
d. 8
e. 10

penyelesaian:
$\lim_{ x\to 2}$$\frac {x^3 - 4x}{x - 2}$ = $\lim_{ x\to 2}$$\frac {x(x^2-4)}{x - 2}$
↔ = $\lim_{ x\to 2}$$\frac {x(x^2-4)}{x - 2}$
↔ = $\lim_{ x\to 2}$$\frac {x(x + 2)(x - 2)}{x - 2}$
↔ = $\lim_{ x\to 2}$ (x (x + 2))
↔ = (2 (2 + 2))
↔ = 2.4
↔ = 8

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3. $\lim_{ x\to 3}$$\frac {x^2 - 9}{x - 3}$ = .....
a. 2
b. 3
c. 4
d. 5
e. 6

penyelesaian:
$\lim_{ x\to 3}$$\frac {x^2 - 9}{x - 3}$ = $\lim_{ x\to 3}$$\frac {(x + 3)(x-3)}{x - 3}$
↔ = $\lim_{ x\to 3}$ (x + 3)
↔ = 3 + 3
↔ = 6

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4. $\lim_{ x\to 2}$$\frac {x^2 - 5x + 6}{x^2 + 2x - 8}$ = .....
a. 3
b. 2½
c. 2
d. 1
e. $\frac {-1}{6}$

penyelesaian:
$\lim_{ x\to 2}$$\frac {x^2 - 5x + 6}{x^2 + 2x - 8}$ = $\lim_{ x\to 1}$$\frac {(x- 2)(x - 3)}{(x - 2)(x + 4)}$
↔ = $\lim_{ x\to 2}$ $\frac {(x - 3)}{(x + 4)}$
↔ = $\frac {(2 - 3)}{(2 + 4)}$
↔ = $\frac {-1}{6}$

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5. $\lim_{ x\to 1}$$\frac {x^2 - 5x + 4}{x^3 - 1}$ = .....
a. 3
b. 2½
c. 2
d. 1
e. -1

penyelesaian:
$\lim_{ x\to 1}$$\frac {x^2 - 5x + 4}{x^3 - 1}$ = $\lim_{ x\to 1}$ $\frac {(x -4)(x - 1)}{(x^2 + x + 1)(x -1)}$
↔ = $\lim_{ x\to 1}$ $\frac {(x -4)}{(x^2 + x + 1)}$
↔ = $\frac {(1 -4)}{(1^2 + 1 + 1)}$
↔ = $\frac {-3}{3}$
↔ = -1

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6. $\lim_{ x\to 3}$ $(\frac {1}{x-3}-\frac {6}{x^2-9})$ = .....
a. $\frac {-1}{6}$
b. $\frac {1}{6}$
c. $\frac {1}{3}$
d. $\frac {1}{2}$
e. 1

penyelesaian:
$\lim_{ x\to 3}$ $(\frac {1}{x-3}-\frac {6}{x^2-9})$ = $\lim_{ x\to 3}$ $(\frac {1}{x-3}-\frac {6}{(x -3)(x+3)})$
↔ = $\lim_{ x\to 3}$ ${(\frac {(x + 3)}{(x -3)(x+3)}-\frac {6}{(x -3)(x+3)})}$
↔ = $\lim_{ x\to 3}$ ${(\frac {(x + 3-6)}{(x -3)(x+3)})}$
↔ = $\lim_{ x\to 3}$ ${(\frac {(x - 3)}{(x -3)(x+3)})}$
↔ = $\lim_{ x\to 3}$ ${(\frac {1}{(x+3)})}$
↔ = $\frac {1}{3+3}$
↔ = $\frac {1}{6}$

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7. $\lim_{ x\to 3}$ ${x^3+ 2x^2 + 5x}$ = .....
a. 20
b. 40
c. 60
d. 70
e. 80

penyelesaian:
$\lim_{ x\to 3}$ ${x^3+ 2x^2 + 5x}$ = ${3^3+ 2.3^2 + 5.3}$
↔ = 27+ 2.9 + 15

↔ = 27+ 18 + 15
↔ = 60

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8. $\lim_{ x\to 1}$$\frac {x^5 - 1}{x - 1}$ = .....
a. 3
b. 4
c. 5
d. 6
e. -1

penyelesaian:
$\lim_{ x\to 1}$$\frac {x^5 - 1}{x - 1}$ = $\lim_{ x\to 1}$ $\frac {(x^4+x^3+x^2+x+1)(x - 1)}{(x -1)}$
↔ = $\lim_{ x\to 1}$ $x^4+x^3+x^2+x+1$
↔ = $1^4+1^3+1^2+1+1$
↔ = 1 + 1 + 1 + 1 + 1
↔ = 5

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9. $\lim_{ x\to 2}$$\frac {x^3 - 4}{x - 4}$ = .....
a. 3
b. 4
c. 5
d. 6
e. 7

penyelesaian:
$\lim_{ x\to 2}$$\frac {x^3 - 4}{x - 4}$ = = $\lim_{ x\to 1}$ $\frac {(x^2+x+1)(x - 4)}{(x - 4)}$
↔ = $\lim_{ x\to 2}$ $x^2+x+1$
↔ = $2^2+ 2 + 1$
↔ = 4 + 3
↔ = 7

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10. $\lim_{ x\to 2}$$\frac {x - 3}{x^2 + x - 12}$ = .....
a. 4
b. 3
c. 3/7
d. 1/7
e. 0

penyelesaian:
$\lim_{ x\to 2}$$\frac {x - 3}{x^2 + x - 12}$ = $\lim_{ x\to 2}$$\frac {x - 3}{(x - 3)(x+4)}$
↔ = $\lim_{ x\to 2}$$\frac {1}{(x+4)}$
↔ = $\frac {1}{(2+4)}$
↔ = $\frac {1}{6}$

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11. Diketahui: $f(x)=\begin{cases}3x-p,\ x\leq 2 \\ 2x+1,\ x \gt 2 \end{cases}$ Agar $\lim\limits_{x \to 2}f(x)$ mempunyai nilai, maka $p=...$
$\begin{align} a.\ & -2 \\ b.\ & -1 \\ c.\ & 0 \\ d.\ & 1 \\ e.\ & 2 \end{align}$

penyelesaian: Berdasarkan definisi limit, agar $\lim\limits_{x \to 2}f(x)$ mempunyai nilai maka Limit Kiri = Limit Kanan secara simbol dituliskan $\lim\limits_{x \to 2^{+}}f(x)=\lim\limits_{x \to 2^{-}}f(x)=L$ Limit kanan $\lim_{x\rightarrow 2^{+}}f(x)$ $\lim\limits_{x \to 2^{+}}(2x+1)=2(2)+1=5$ Limit kiri $\lim_{x\rightarrow 2^{-}}f(x)$ $\lim\limits_{x \to 2^{-}}(3x-p)=3(2)-p=6-p$ Berdasarkan definisi agar $\lim\limits_{x \to 2}f(x)$ mempunyai nilai yaitu Limit Kiri = Limit Kanan maka: $6-p=5$ $6-5=p$ $p=1$
● Pilihan yang sesuai adalah $d.\ 1$

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12. Nilai $\lim\limits_{x \to 2} \dfrac{2x^{2}-x-6}{3x^{2}-5x-2} =\cdots$
$\begin{align} a.\ & -1 \\ b.\ & 0 \\ c.\ & \dfrac{1}{5} \\ d.\ & 1 \\ e.\ & 7 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 2} \dfrac{2x^{2}-x-6}{3x^{2}-5x-2} \\ & = \lim\limits_{x \to 2} \dfrac{\left( 2x+3 \right)\left( x-2 \right)}{\left( 3x+1 \right)\left( x-2 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{\left( 2x+3 \right)}{\left( 3x+1 \right)} \\ & = \dfrac{2(2)+3 }{3(2)+1}=\dfrac{7}{7}=1 \end{align}$
● Pilihan yang sesuai adalah $d.\ 1$

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13. $\lim\limits_{x \to 3} \dfrac{x^{2}-9}{2x^{2}-7x+3} =\cdots$
$\begin{align} a.\ & \dfrac{1}{2} \\ b.\ & \dfrac{5}{6} \\ c.\ & \dfrac{6}{7} \\ d.\ & \dfrac{7} {6} \\ e.\ & \dfrac{6}{5} \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 3} \dfrac{x^{2}-9}{2x^{2}-7x+3} \\ & = \lim\limits_{x \to 3} \dfrac{(x+3)(x-3)}{(2x-1)(x-3)} \\ & = \lim\limits_{x \to 3} \dfrac{(x+3) }{(2x-1) } \\ & = \dfrac{(3+3) } {(2(3)-1) } = \dfrac{6}{5} \end{align}$
● Pilihan yang sesuai adalah $e.\ \dfrac{6}{5}$

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14. Nilai $\lim\limits_{x \to 3} \dfrac{x-3}{x^{2}+x-12} =\cdots$
$\begin{align} a.\ & 4 \\ b.\ & 3 \\ c.\ & \dfrac{3}{7} \\ d.\ & \dfrac{1}{7} \\ e.\ & 0 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 3} \dfrac{x-3}{x^{2}+x-12} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x-3 \right)}{ \left( x+4 \right)\left( x-3 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{ \left( 1 \right)} {\left( x+4 \right)} \\ & = \dfrac{ 1 }{3+4}=\dfrac{ 1 }{7} \end{align}$
● Pilihan yang sesuai adalah $d.\ \dfrac{1}{7}$

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15. Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}-x-2} =\cdots$
$\begin{align} a.\ & 3 \\ b.\ & 2 \\ c.\ & 0 \\ d.\ & -2 \\ e.\ & -3 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}-x-2} \\ & = \lim\limits_{x \to 2} \dfrac{\left( x+4 \right)\left( x-2 \right)}{ \left( x-2 \right)\left( x+1 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( x+4 \right)}{\left( x+1 \right)} \\ & = \dfrac{ 2+4 }{2+1}=\dfrac{ 6 }{3}=2 \end{align}$
● Pilihan yang sesuai adalah $b.\ 2$

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16. Nilai $\lim\limits_{x \to 3} \dfrac{\left( x-2 \right)^{2}-1}{x-3} =\cdots$
$\begin{align} a.\ & 0 \\ b.\ & 1 \\ c.\ & 2 \\ d.\ & 4 \\ e.\ & 6 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 3} \dfrac{\left( x-2 \right)^{2}-1}{x-3} \\ & = \lim\limits_{x \to 3} \dfrac{x^{2}-4x+4-1}{x-3} \\ & = \lim\limits_{x \to 3} \dfrac{x^{2}-4x+3}{x-3} \\ & = \lim\limits_{x \to 3} \dfrac{\left( x-1 \right)\left( x-3 \right)}{ \left( x-3 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{ \left( x-1 \right)}{\left( 1 \right)} \\ & = \dfrac{ 3-1 }{1}=2 \end{align}$
● Pilihan yang sesuai adalah $c.\ 2$

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17. Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}-5x+6}{x^{2}-4} =\cdots$
$\begin{align} a.\ & -\dfrac{1}{4} \\ b.\ & -\dfrac{1}{8} \\ c.\ & \dfrac{1}{8} \\ d.\ & 1 \\ e.\ & \dfrac{5}{4} \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}-5x+6}{x^{2}-4} \\ & = \lim\limits_{x \to 2} \dfrac{\left( x-2 \right)\left( x-3 \right)}{ \left( x+2 \right)\left( x-2 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( x-3 \right)}{\left( x+2 \right)} \\ & = \dfrac{ 2-3 }{2+2}=\dfrac{ -1 }{4} \end{align}$
● Pilihan yang sesuai adalah $a.\ -\dfrac{1}{4}$

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18. Nilai $\lim\limits_{x \to 2} \dfrac{x^{3}-4x}{x-2} =\cdots$
$\begin{align} a.\ & 32 \\ b.\ & 16 \\ c.\ & 8 \\ d.\ & 4 \\ e.\ & 2 \end{align}$

penyelesaian: Kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 2} \dfrac{x^{3}-4x}{x-2} \\ & = \lim\limits_{x \to 2} \dfrac{ x\ \left( x-2 \right)\left( x+2 \right)}{x-2} \\ & = \lim\limits_{x \to 2} \dfrac{ x\ \left( x+2 \right)}{1} \\ & = \dfrac{2\left( 2+2 \right)}{1}=8 \end{align}$
● Pilihan yang sesuai adalah $c.\ 8$

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19. Nilai $\lim\limits_{x \to 1} \dfrac{x^{2}-5x+4}{x^{3}-1} =\cdots$
$\begin{align} a.\ & 3 \\ b.\ & 2\frac{1}{2} \\ c.\ & 2 \\ d.\ & 1 \\ e.\ & -1 \end{align}$

penyelesaian: Kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 1} \dfrac{x^{2}-5x+4}{x^{3}-1} \\ & = \lim\limits_{x \to 1} \dfrac{ \left( x-1 \right)\left( x-4 \right)}{\left( x-1 \right)\left( x^{2}+x+1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{ \left( x-4 \right)}{ \left( x^{2}+x+1 \right)} \\ & = \dfrac{ \left( 1-4 \right)}{ \left( (1)^{2}+(1)+1 \right)} \\ & = \dfrac{-3}{3} =-1 \end{align}$
● Pilihan yang sesuai adalah $e.\ -1$

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20. Jika $p \gt 0$ dan $\lim\limits_{x \to p} \dfrac{ x^{3}+px^{2}+qx}{x-p}=12$, maka nilai $p-q$ adalah...
$\begin{align} a.\ & 14 \\ b.\ & 10 \\ c.\ & 8 \\ d.\ & 5 \\ e.\ & 3 \end{align}$

penyelesaian: Nilai $\lim\limits_{x \to p} \dfrac{ x^{3}+px^{2}+qx}{x-p}=12$ sehingga jika kita substitusi langsung nilai $x=p$ maka nilai $x^{3}+px^{2}+qx$ harus $0$, karena jika $x^{3}+px^{2}+qx$ tidak nol maka nilai limit adalah $\infty$. Karena nilai $x^{3}+px^{2}+qx$ untuk $x=p$ adalah $0$ maka dapat kita tuliskan:
$\begin{align} (p)^{3}+p(p)^{2}+q(p) &= 0 \\ 2p^{3} + pq &= 0 \\ 2p^{2} + q &= 0 \\ q &= 2p^{2} \\ \end{align}$ $x-p$ adalah salah satu faktor $x^{3}+px^{2}+qx$ sehingga dapat kita tuliskan;
$\begin{align} x^{3}+px^{2}+qx &= (x-p)( x^{2}+bx+c) \\ x^{3}+px^{2}+qx &= x^{3}+bx^{2}+cxpx^{ 2}-bpx-pc \\ x^{3}+px^{2}+qx &= x^{3}+ \left( b-p \right)x^{2}+\left( c-bp \right)x-pc \\ \hline - pc=0 & \rightarrow c=0 \\ b-p=p & \rightarrow b=2p \\ \hline x^{3}+px^{2}+qx &= (x-p)( x^{2}+2px) \\ \end{align}$ Dari hasil di atas kita peroleh:
$\begin{align} \lim\limits_{x \to p} \dfrac{ x^{3}+px^{2}+qx }{x-p} &=12 \\ \lim\limits_{x \to p} \dfrac{ (x-p)( x^{2}+2px) }{x-p} &=12 \\ \lim\limits_{x \to p} \dfrac{ ( x^{2}+2px) }{1} &=12 \\ (p)^{2}+2p(p) &=12 \\ 3p^{2} &=12 \\ p^{2} &= 4 \rightarrow p=2 \\ \hline q &= -2p^{2} \rightarrow q=-8 \end{align}$

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21. Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}-3x+2}{x-1} =\cdots$
$\begin{align} a.\ & -2 \\ b.\ & -1 \\ c.\ & 0 \\ d.\ & 1 \\ e.\ & 2 \end{align}$

penyelesaian: Kita kerjakan langsung dengan substitusi,
$\begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}-3x+2}{x-1} \\ & = \dfrac{2^{2}-3(2)+2}{2-1} \\& = \dfrac{4-6+2}{2-1} \\ & = \dfrac{ 0 }{1}= 0 \end{align}$
● Pilihan yang sesuai adalah $c.\ 0$

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22. Nilai dari $\lim\limits_{x \to 0} \dfrac{6x^{5}-4x}{2x^{4}+x} =\cdots$
$\begin{align} a.\ & -4 \\ b.\ & -2 \\ c.\ & 0 \\ d.\ & 2 \\ e.\ & 4 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 0} \dfrac{6x^{5}-4x}{2x^{4}+x} \\ & = \lim\limits_{x \to 0} \dfrac{\left( x \right) \left( 6x^{4}-4 \right)}{\left( x \right) \left( 2x^{3}+1 \right)} \\ & = \lim\limits_{x \to 0} \dfrac{ \left( 6x^{4}-4 \right)}{\left( 2x^{3}+1 \right)} \\ & = \dfrac{ \left( 6(0)^{4}-4 \right)}{\left( 2(0)^{3}+1 \right)} \\ & = \dfrac{ -4 }{1}=-4 \end{align}$
● Pilihan yang sesuai adalah $a.\ -4$

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23. Nilai $\lim\limits_{x \to 5} \dfrac{x^{2}-x-20}{x-5} =\cdots$
$\begin{align} a.\ & 9 \\ b.\ & 5 \\ c.\ & 4 \\ d.\ & -4 \\ e.\ & -9 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 5} \dfrac{x^{2}-x-20}{x-5} \\ & = \lim\limits_{x \to 5} \dfrac{\left( x-5 \right) \left( x+4 \right)}{\left( x-5 \right)} \\ & = \lim\limits_{x \to 5} \dfrac{ \left( x+4 \right)} {\left( 1 \right)} \\ & = \dfrac{ 5+4 }{1}=9 \end{align}$
● Pilihan yang sesuai adalah $a.\ 9$

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24. Nilai $\lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}+4x-12} =\cdots$
$\begin{align} a.\ & \infty \\ b.\ & 1 \\ c.\ & \dfrac{3}{4} \\ d.\ & \dfrac{1}{2} \\ e.\ & 0 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 2} \dfrac{x^{2}+2x-8}{x^{2}+4x-12} \\ & = \lim\limits_{x \to 2} \dfrac{\left( x+4 \right)\left( x-2 \right)}{ \left( x+6 \right)\left( x-2 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{ \left( x+4 \right)}{\left( x+6 \right)} \\ & = \dfrac{ 2+4 }{2+6}=\dfrac{ 6 }{8}=\dfrac{ 3 }{4} \end{align}$
● Pilihan yang sesuai adalah $c.\ \dfrac{3}{4}$

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25. Nilai $\lim\limits_{x \to 3} \dfrac{2x^{2}-4x-6}{x^{2}-2x-3} =\cdots$
$\begin{align} a.\ & -2 \\ b.\ & 0 \\ c.\ & 2 \\ d.\ & 6 \\ e.\ & 8 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 3} \dfrac{2x^{2}-4x-6}{x^{2}-2x-3} \\ & = \lim\limits_{x \to 3} \dfrac{2 \left( x+1 \right)\left( x-3 \right)}{ \left( x+1 \right)\left( x-3 \right)} \\ & = \lim\limits_{x \to 3} \dfrac{2 \left( x+1 \right)}{ \left( x+1 \right)} \\ & = \dfrac{ 2(3+1) }{3+1}=2 \end{align}$
● Pilihan yang sesuai adalah $c.\ 2$

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26. Nilai $\lim\limits_{x \to -4} \dfrac{x^{2}+7x+12}{2x+8} =\cdots$
$\begin{align} a.\ & -1 \\ b.\ & -\dfrac{1}{2} \\ c.\ & \dfrac{7}{8} \\ d.\ & \dfrac{3}{2} \\ e.\ & \dfrac{7}{2} \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to -4} \dfrac{x^{2}+7x+12}{2x+8} \\ & = \lim\limits_{x \to -4} \dfrac{\left( x+3 \right)\left( x+4 \right)}{ \left( 2 \right)\left( x+4 \right)} \\ & = \lim\limits_{x \to -4} \dfrac{ \left( x+3 \right)}{\left( 2 \right)} \\ & = \dfrac{ -4+3 }{2}=\dfrac{ -1 }{2} \end{align}$
● Pilihan yang sesuai adalah $b.\ -\dfrac{1}{2}$

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27. Nilai $\lim\limits_{x \to 4} \dfrac{x^{2}-16}{x-4} =\cdots$
$\begin{align} a.\ & 16 \\ b.\ & 8 \\ c.\ & 4 \\ d.\ & -4 \\ e.\ & -8 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 4} \dfrac{x^{2}-16}{x-4} \\ & = \lim\limits_{x \to 4} \dfrac{\left( x+4 \right)\left( x-4 \right)}{ \left( x-4 \right)} \\ & = \lim\limits_{x \to 4} \dfrac{ \left( x+4 \right)} {\left( 1 \right)} \\ & = \dfrac{ 4+4 }{1}=8 \end{align}$
● Pilihan yang sesuai adalah $b.\ 8$

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28. Nilai $\lim\limits_{x \to 2} \left( \dfrac{2}{x-2}-\dfrac{8}{x^{2}-4} \right)=\cdots$
$\begin{align} a.\ & \dfrac{1}{4} \\ b.\ & \dfrac{1}{2} \\ c.\ & 2 \\ d.\ & 4 \\ e.\ & \infty \end{align}$

penyelesaian: Kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 2} \left( \dfrac{2}{x-2}-\dfrac{8}{x^{2}-4} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{\left( 2x^{2}-8 \right)-\left( 8x-16 \right)}{\left( x-2 \right)\left( x^{2}-4 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{2x^{2}-8x+8}{\left( x-2 \right)\left( x-2 \right)\left( x+2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{2\left( x-2 \right)\left( x-2 \right)}{\left( x-2 \right)\left( x-2 \right)\left( x+2 \right)\left( x-2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{2}{\left( x+2 \right)} \right) \\ & = \dfrac{2}{\left( 2+2 \right)}=\dfrac{2}{4} \end{align}$
● Pilihan yang sesuai adalah $b.\ \dfrac{1}{2}$

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29. $\lim\limits_{x \to 2} \dfrac{x^{3}-8}{x^{2}+x-6} =\cdots$
$\begin{align} a.\ & \dfrac{3}{4} \\ b.\ & \dfrac{2}{15} \\ c.\ & 1 \dfrac{1}{3} \\ d.\ & 2 \dfrac{2}{5} \\ e.\ & 6 \end{align}$

penyelesaian: Pertama kita kerjakan dengan cara memfaktorkan,
$\begin{align} & \lim\limits_{x \to 2} \dfrac{x^{3}-8}{x^{2}+x-6} \\ & = \lim\limits_{x \to 2} \dfrac{\left(x^{2}+2x+4 \right)\left( x-2 \right)}{\left( x-2 \right)\left( x+3 \right)} \\ & = \lim\limits_{x \to 2} \dfrac{\left(x^{2}+2x+4 \right)}{\left( x+3 \right)} \\ & = \dfrac{ (2)^{2}+2(2)+4} {2+3} = \dfrac{ 12}{5} \end{align}$
● Pilihan yang sesuai adalah $d.\ 2 \dfrac{2}{5}$

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30. Nilai $\lim\limits_{x \to 2} \left( \dfrac{2}{x^{2}-4}-\dfrac{3}{x^{2}+2x-8} \right)=\cdots$
$\begin{align} a.\ & -\dfrac{7}{12} \\ b.\ & -\dfrac{1}{4} \\ c.\ & -\dfrac{1}{12} \\ d.\ & - \dfrac{1}{24} \\ e.\ & 0 \end{align}$

penyelesaian:
$\begin{align} & \lim\limits_{x \to 2} \left( \dfrac{2}{x^{2}-4}-\dfrac{3}{x^{2}+2x-8} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{\left( 2x^{2}+4x-16 \right)-\left( 3x^{2}-12 \right)}{\left( x^{2}+2x-8 \right)\left( x^{2}-4 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{-x^{2}+4x-4}{\left( x+2 \right)\left( x-2 \right)\left( x+4 \right)\left( x-2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{-\left( x-2 \right)\left( x-2 \right)}{\left( x+2 \right)\left( x-2 \right)\left( x+4 \right)\left( x-2 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{-1}{\left( x+2 \right) \left( x+4 \right)} \right) \\ & = \dfrac{-1}{\left( 2+2 \right) \left( 2+4 \right)}=\dfrac{-1}{24} \end{align}$
● Pilihan yang sesuai adalah $a.\ -\dfrac{1}{24}$

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31. Jika $\lim\limits_{x \to -3} \dfrac{\frac{1}{ax}+\frac{1}{3}}{bx^{3}+27}=-\dfrac{1}{3^{5}}$, nilai $a+b$ untuk $a$ dan $b$ bulat positif adalah...
$\begin{align} a.\ & -4 \\ b.\ & -2 \\ c.\ & 0 \\ d.\ & 2 \\ e.\ & 4 \end{align}$

penyelesaian: menggunakan metode turunan pada limit untuk $x \to -3$ hasilnya adalah $-\dfrac{1}{3^{5}}$.
$\begin{align} \lim\limits_{x \to -3} \dfrac{\frac{1}{ax}+\frac{1}{3}}{bx^{3}+27} & = -\dfrac{1}{3^{5}} \\ \lim\limits_{x \to -3} \dfrac{\frac{1}{a}x^{-1}+\frac{1}{3}}{bx^{3}+27} & = -\dfrac{1}{3^{5}} \\ \lim\limits_{x \to -3} \dfrac{-\frac{1}{ax^{2}}}{3bx^{2}} & = -\dfrac{1}{3^{5}} \\ \lim\limits_{x \to -3} \dfrac{-1}{3bx^{2} \cdot ax^{2}} & = -\dfrac{1}{3^{5}} \\ \lim\limits_{x \to -3} \dfrac{-1}{3abx^{4}} & = -\dfrac{1}{3^{5}} \\ -\dfrac{1}{3ab(-3)^{4}} & = -\dfrac{1}{3^{5}} \\ -\dfrac{1}{ ab3^{5}} & = - \dfrac{1}{3^{5}} \\ ab & = 1 \end{align}$ Untuk $a$ dan $b$ bilangan bulat positif yang memenuhi $ab=1$ adalah $a=1$ dan $b=1$, maka $a+b=2$

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32. $\lim\limits_{x \to 1} \left( \dfrac{1}{1-x}-\dfrac{2}{x-x^{3}} \right)=\cdot$
$\begin{align} a.\ & -\dfrac{3}{2} \\ b.\ & -\dfrac{2}{3} \\ c.\ & \dfrac{2}{3} \\ d.\ & 1 \\ e.\ & \dfrac{3}{2} \end{align}$

penyelesaian:
$\begin{align} & \lim\limits_{x \to 1} \left( \dfrac{1}{1-x}-\dfrac{2}{x-x^{3}} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{1}{1-x}-\dfrac{2}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{\left( x \right)\left( 1+x \right)-2}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{x^{2}+x-2}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{ \left( x+2 \right)\left( x-1 \right)}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x\to 1} \left( \dfrac{ -\left( x+2 \right)\left( 1-x \right)}{\left( x \right)\left( 1-x \right)\left( 1+x \right)} \right) \\ & = \lim\limits_{x \to 1} \left( \dfrac{ -\left( x+2 \right)}{\left( x \right)\left( 1+x \right)} \right) \\ & = \dfrac{-\left( (1)+2 \right)}{\left( 1 \right)\left( 1+(1) \right)} \\ & = \dfrac{-3} {2} \end{align}$
● Pilihan yang sesuai adalah $a.\ -\dfrac{3}{2}$

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33. Diketahui $f(x)=5x^{2}+3$. Hasil dari $\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ adalah...
$\begin{align} a.\ & 0 \\ b.\ & 5 \\ c.\ & 10 \\ d.\ & 10x \\ e.\ & 5x^{2} \end{align}$

penyelesaian: Dari informasi pada soal, yang ditanyakan adalah $\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ dari sebuah fungsi $f(x)$. Jika kita teliti dalam membaca soal bahwa yang ditanyakan pada soal adalah turunan fungsi $f(x)$. Definisi turunan fungsi $f(x)$ adalah $f'(x)=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.
$\begin{align} f(x) &=5x^{2}+3 \\ f'(x) &=10x \end{align}$ Tetapi jika ingin mengerjakannya dengan proses limit fungsi berikut pembahasannya:
$\begin{align} & \lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h} \\ & = \lim\limits_{h \to 0} \dfrac{ 5(x+h)^{2}+3 -\left( 5x^{2}+3 \right)}{h} \\ & = \lim\limits_{h \to 0} \dfrac{ 5 \left(x^{2}+2hx+h^{2} \right) +3 - 5x^{2}-3}{h} \\ & = \lim\limits_{h \to 0} \dfrac{ 5 x^{2}+10hx+5h^{2} +3 - 5x^{2}-3}{h} \\ & = \lim\limits_{h \to 0} \dfrac{ +10hx+5h^{2}}{h} \\ & = \lim\limits_{h \to 0} \left( +10 x+5h \right) \\ & = 10 x+5(0) \\ & = 10x \end{align}$
● Pilihan yang sesuai adalah $d.\ 10x$

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34. $\lim\limits_{t \to 2} \dfrac{4t^{4}+4t-72}{\left( t-2 \right)\left( t^{2}+3t+2 \right)}=\cdot$
$\begin{align} a.\ & \dfrac{11}{4} \\ b.\ & \dfrac{11}{3} \\ c.\ & 11 \\ d.\ & 22 \\ e.\ & 33 \end{align}$

penyelesaian:
$\begin{align} & \lim\limits_{t \to 2} \dfrac{4t^{4}+4t-72}{\left( t-2 \right)\left( t^{2}+3t+2 \right)} \\ & = \lim\limits_{t \to 2} \dfrac{4\left( t-2 \right)\left( t^{3}+2t^{2}+4t+9 \right)}{\left( t-2 \right)\left( t^{2}+3t+2 \right)} \\ & = \lim\limits_{t \to 2} \dfrac{4 \left( t^{3}+2t^{2}+4t+9 \right)}{ \left( t^{2}+3t+2 \right)} \\ & = \dfrac{4 \left( (2)^{3}+2(2)^{2}+4(2)+9 \right)}{ (2)^{2}+3(2)+2} \\ & = \dfrac{4 \left( 33 \right)}{ 12} = 11 \end{align}$
● Pilihan yang sesuai adalah $c.\ 11$

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35. Nilai $\lim\limits_{x \to 2} \left( \dfrac{6}{x^{2}-x-2}-\dfrac{2}{x-2} \right)$ sama dengan...
$\begin{align} a.\ & -1 \\ b.\ & -\dfrac{2}{3} \\ c.\ & -\dfrac{1}{3} \\ d.\ & \dfrac{1}{3} \\ e.\ & \dfrac{2}{3} \end{align}$

penyelesaian:
$\begin{align} & \lim\limits_{x \to 2} \left( \dfrac{6}{x^{2}-x-2}-\dfrac{2}{x-2} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{6}{\left( x-2 \right)\left( x+1 \right)}-\dfrac{2}{x-2} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{6}{\left( x-2 \right)\left( x+1 \right)}-\dfrac{2\left( x+1 \right)}{\left( x-2 \right)\left( x+1 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{6-2\left( x+1 \right)}{\left( x-2 \right)\left( x+1 \right)} \right)\\ & = \lim\limits_{x \to 2} \left( \dfrac{4-2x}{\left( x-2 \right)\left( x+1 \right)} \right) \\ & = \lim\limits_{x \to 2} \left( \dfrac{-2\left( x-2 \right)}{\left( x-2 \right)\left( x+1 \right)} \right)\\ & = \lim\limits_{x \to 2} \left( \dfrac{-2 }{ \left( x+1 \right)} \right)\\ & = \dfrac{-2 }{ \left( 2+1 \right)} = \dfrac{-2}{3} \end{align}$
● Pilihan yang sesuai adalah $b.\ -\dfrac{2}{3}$

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36. Jika $\lim\limits_{x \to a} \left( f(x)-3g(x) \right)=2$ dan $\lim\limits_{x \to a} \left( 3f(x)+g(x) \right)=1$ maka $\lim\limits_{x \to a} \left( f(x) \cdot g(x) \right)=\cdots$
$\begin{align} a.\ & -\dfrac{1}{2} \\ b.\ & -\dfrac{1}{4} \\ c.\ & \dfrac{1}{4} \\ d.\ & \dfrac{1}{2} \\ e.\ & 1 \end{align}$

penyelesaian: Dari persamaan $\lim\limits_{x \to a} \left( f(x)-3g(x) \right)=2$ dan $\lim\limits_{x \to a} \left( 3f(x)+g(x) \right)=1$ dengan teorema limit dapat kita ubah betuknya menjadi:
$\begin{align} \lim\limits_{x \to a} \left( f(x)-3g(x) \right) &=2 \\ \lim\limits_{x \to a} f(x)-3\ \lim\limits_{x \to a} g(x) &=2 \\ \hline \lim\limits_{x \to a} \left( 3f(x)+g(x) \right) &=1 \\ 3\ \lim\limits_{x \to a} f(x)+ \lim\limits_{x \to a} g(x) &=1 \end{align}$ Jika kita misalkan $\lim\limits_{x \to a} f(x)=m$ dan $\lim\limits_{x \to a} f(x)=n$, maka kita peroleh: $\begin{array}{c|c|cc} m-3n=2 & (\times 3) \\ 3m+ n =1 & (\times 1) \\ \hline 3m - 9n =6 & \\ 3m + n =1 \ \ \ (-)& \\ \hline -10n = 5 & \\ n = -\dfrac{1}{2} & \lim\limits_{x \to a} g(x)=-\dfrac{1}{2} \\ m = \dfrac{1}{2} & \lim\limits_{x \to a} f(x) = \dfrac{1}{2} \\ \end{array}$
$\begin{align} \lim\limits_{x \to a} \left( f(x) \cdot g(x) \right) & = \lim\limits_{x \to a} f(x) \cdot \lim\limits_{x \to a} g(x) \\ & = \left( -\dfrac{1}{2} \right) \cdot \left( \dfrac{1}{2} \right) \\ & = - \dfrac{1}{4} \end{align}$
● Pilihan yang sesuai adalah $b.\ -\dfrac{1}{4}$

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37. Jika $a$ dan $b$ adalah dua bilangan real dengan $\lim\limits_{x \to 2} \dfrac{x^{2}+2ax+b} {x-2}=-3$, maka $ab=\cdots$
$\begin{align} a.\ & -35 \\ b.\ & -30 \\ c.\ & -15 \\ d.\ & -3 \\ e.\ & -1 \end{align}$

penyelesaian: Nilai $\lim\limits_{x \to 2} \dfrac{ x^{2}+2ax+b}{x-2}=-3$ sehingga jika kita substitusi langsung nilai $x=2$ maka nilai $x^{2}+2ax+b$ harus $0$, karena jika $x^{2}+2ax+b$ tidak nol maka nilai limit adalah $\infty$. Karena nilai $x^{2}+2ax+b$ untuk $x=2$ adalh $0$ maka $x-2$ adalah salah satu faktornya sehingga $x^{2}+2ax+b \equiv (x-2)( x+n)$, dan dapat kita tuliskan;
$\begin{align} \lim\limits_{x \to 2} \dfrac{ x^{2}+2ax+b}{x-2} &=-3 \\ \lim\limits_{x \to 2} \dfrac{ (x- 2)( x+n)}{x-2} &=-3 \\ \lim\limits_{x \to 2} \left( x+n \right) &=-3 \\ 2+n &=-3 \\ n &=-5 \end{align}$ Untuk $n=-5$, kita peroleh:
$\begin{align} x^{2}+2ax+b &= \equiv (x-2)( x+n) \\ x^{2}+2ax+b &= \equiv (x-2)( x-5) \\ x^{2}+2ax+b &= \equiv x^{2}-7x+10 \\ \hline 2a &=-7 \\ b &=10 \\ ab &= -35 \end{align}$
● Pilihan yang sesuai $a.\ -35$

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38. Jika kurva $f(x)=ax^{2}+bx+c$ memotong sumbu-$y$ di titik $(0,1)$ dan $\lim\limits_{x \to 1} \dfrac{f(x)}{x-1}=-4$ maka $\dfrac{b+c}{a}=\cdots$
$\begin{align} a.\ & -1 \\ b.\ & -\dfrac{1}{2} \\ c.\ & 0 \\ d.\ & 1 \\ e.\ & \dfrac{3}{2} \end{align}$

penyelesaian: Kurva $f(x)=ax^{2}+bx+c$ memotong sumbu-$y$ di titik $(0,1)$ maka nilai $c=1$ sehingga $f(x)=ax^{2}+bx+1$. Nilai $\lim\limits_{x \to 1} \dfrac{ax^{2}+bx+1}{x-1}=-4$ Jika kita substitusi langsung nilai $x=1$ maka nilai $ax^{2}+bx+1$ harus $0$, karena jika $ax^{2}+bx+1$ tidak nol maka nilai limit adalah $\infty$. Karena nilai $ax^{2}+bx+1$ untuk $x=1$ adalh $0$ maka $x-1$ adalah salah satu faktornya sehingga berlaku;
$\begin{align} ax^{2}+bx+1 & \equiv (x-1)(mx+n) \\ ax^{2}+bx+1 & \equiv mx^{2}+nx-mx-n \\ ax^{2}+bx+1 & \equiv mx^{2}+(n-m)x-n \\ -1 &= n \\ b &= n-m \\ b &= -1-m \\ a &= m \end{align}$ Nilai $\lim\limits_{x \to 1} \dfrac{ax^{2}+bx+1}{x-1}=-4$, maka:
$\begin{align} \lim\limits_{x \to 1} \dfrac{(x-1)(mx+n)}{x-1} & =-4 \\ \lim\limits_{x \to 1} (mx+n) & =-4 \\ \lim\limits_{x \to 1} (mx-1) & =-4 \\ m-1 & =-4 \\ m &= -4+1 \\ m &=-3 \end{align}$ Untuk $m=-3$ nilai $a=-3$, $b=2$ dan $c=1$, maka $\dfrac{b+c}{a}=\dfrac{2+1}{-3}=-1$
● Pilihan yang sesuai $a.\ -1$

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39. $\lim\limits_{x \to 1} \dfrac{x^{2n}- x}{1-x}=\cdots$
$\begin{align} a.\ & 2n-1 \\ b.\ & 1-2n \\ c.\ & 2n \\ d.\ & 2n-2 \\ e.\ & 2n+2 \end{align}$

penyelesaian: Beberapa sifat aljabar yang mungkin dapat membantu dalam menyelesaikan limit fungsi di atas, antara laia: $a^{n}-b^{n}=\left(a-b \right)\left(a^{n-1}+a^{n-2}b+ \cdots + ab^{n-2}+b^{n-1} \right)$ Untuk $n$ bilangan Asli
$\begin{align} & \lim\limits_{x \to 1} \dfrac{x^{2n}- x}{1-x} \\ & = \lim\limits_{x \to 1} \dfrac{x\left( x^{2n-1}- 1 \right)}{-\left( x -1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{x \left( x^{2n-1}- 1^{2n-1} \right) }{-\left( x -1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{x \left(x-1 \right)\left(x^{2n-1- 1}+x^{2n-1-2}(1)+ \cdots + (x)(1)^{2n-1-2}+(1)^{2n-1-1} \right) }{-\left( x -1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{x \left(x-1 \right)\left(x^{2n-2}+x^{2n-3} + \cdots + x+1 \right) }{-\left( x -1 \right)} \\ & = \lim\limits_{x \to 1} \dfrac{x \left(x^{2n-2}+x^{2n-3} + \cdots + x+1 \right) }{-1} \\ & = \dfrac{(1) \left((1)^{2n-2}+(1)^{2n-3} + \cdots + (1)+1 \right) }{-1} \\ & = -(1) \left( 2n-2 +1 \right) \\ & = -(1) \left( 2n-1 \right) \\ & = -2n+1 \end{align}$
● Pilihan yang sesuai adalah $b.\ 1-2n$

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40. Diketahui $f(x)=x^{2}+ax+b$ dengan $f \left( b+1 \right)=0$ dan $\lim\limits_{x \to 0} \dfrac{ f \left( x+b \right)}{x}=-1$, maka $a+2b=\cdots$
$\begin{align} a.\ & -2 \\ b.\ & -1 \\ c.\ & 0 \\ d.\ & 1 \\ e.\ & 2 \end{align}$

penyelesaian: Diketahui $f(b+1)=0$ dan $f(x)=x^{2}+ax+b$ sehingga kita peroleh:
$\begin{align} f \left( b+1 \right)\ & = \left( b+1 \right)^{2}+a\left( b+1 \right)+b \\ 0\ & = b^{2}+2b+1+ab+a+b \\ 0\ & = b^{2}+3b+ab+a+1 \end{align}$ Nilai $\lim\limits_{x \to 0} \dfrac{ f \left( x+b \right)}{x}=-1$ sehingga jika kita substitusi langsung nilai $x=0$ maka nilai $f \left( 0+b \right)$ harus $0$. Karena jika $f \left( 0+b \right)$ tidak nol maka nilai limit adalah $\infty$. Karena nilai $f \left(b \right)$ untuk $x=0$ adalah $0$ maka dapat kita tuliskan:
$\begin{align} f(x) & = x^{2}+ax+b \\ f(b) & = (b)^{2}+a(b)+b \\ 0 & = b^{2}+ab+b \\ \hline 0\ & = b^{2}+2b+ab+a+b+1 \\ 0\ & = b^{2}+ab+b+2b+a+1 \\ 0\ & = 0+2b+a+1 \\ -1\ & = 2b+a \end{align}$
● Pilihan yang sesuai $b.\ -1$

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41. Jika $f(x)=x^{2}+ax+b$ dengan $f(2)=0$ dan $\lim\limits_{x \to 2} \dfrac{ f \left( x+1 \right)-f \left( x \right)}{x-2}=2$, maka $b=\cdots$
$\begin{align} a.\ & -6 \\ b.\ & -5 \\ c.\ & 0 \\ d.\ & 5 \\ e.\ & 6 \end{align}$

penyelesaian: Diketahui $f(2)=0$ dan $f(x)=x^{2}+ax+b$ sehingga kita peroleh:
$\begin{align} f(2)\ & = 2^{2}+a(2)+b \\ 0\ & = 4+2a +b \\ -4\ & = 2a +b \end{align}$ Nilai $\lim\limits_{x \to 2} \dfrac{ f \left( x+1 \right)-f \left( x \right)}{x-1}=2$ sehingga jika kita substitusi langsung nilai $x=2$ maka nilai $f \left( x+1 \right)-f \left( x \right)$ harus $0$. Karena jika $f \left( x+1 \right)-f \left( x \right)$ tidak nol maka nilai limit adalah $\infty$. Karena nilai $f \left( x+1 \right)-f \left( x \right)$ untuk $x=2$ adalah $0$ maka dapat kita tuliskan:
$\begin{align} f \left( x+1 \right)-f \left( x \right) & =0 \\ f \left( 2+1 \right)-f \left( 2 \right) & =0 \\ f \left( 3 \right)- 0 & =0 \\ 3^{2}+a(3)+b & =0 \\ 9+3a+b & =0 \\ \hline 3a+b & = -9 \\ 2a +b & = -4\ (-)\\ \hline a & = -5 \\ b & = 6 \end{align}$
● Pilihan yang sesuai $e.\ 6$

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42. Jika $f(x)=x^{2}+ax+b$ dengan $f(1)=0$ dan $\lim\limits_{x \to 1} \dfrac{ f \left( x+1 \right)-f \left( x \right)}{x-1}=2$, maka $b=\cdots$
$\begin{align} a.\ & -2 \\ b.\ & -1 \\ c.\ & 0 \\ d.\ & 1 \\ e.\ & 2 \end{align}$

penyelesaian: Diketahui $f(1)=0$ dan $f(x)=x^{2}+ax+b$ sehingga kita peroleh:
$\begin{align} f(1)\ & = 1^{2}+a(1)+b \\ 0\ & = 1+a +b \\ -1\ & = a +b \end{align}$ Nilai $\lim\limits_{x \to 1} \dfrac{ f \left( x+1 \right)-f \left( x \right)}{x-1}=2$ sehingga jika kita substitusi langsung nilai $x=1$ maka nilai $f \left( x+1 \right)-f \left( x \right)$ harus $0$, karena jika $f \left( x+1 \right)-f \left( x \right)$ tidak nol maka nilai limit adalah $\infty$. Nilai $f \left( x+1 \right)-f \left( x \right)$ untuk $x=1$ adalah $0$ sehingga dapat kita tuliskan:
$\begin{align} f \left( x+1 \right)-f \left( x \right) & =0 \\ f \left( 1+1 \right)-f \left( 1 \right) & =0 \\ f \left( 2 \right)- 0 & =0 \\ 2^{2}+a(2)+b & =0 \\ 4+2a+b & =0 \\ \hline 2a+b & = -4 \\ a +b & = -1\ (-) \\ \hline a & = -3 \\ b & = 2 \end{align}$
● Pilihan yang sesuai $e.\ 2$ \end{align}$\end{align}$