SBMPTN Zone : Limit

Berikut ini adalah kumpulan soal dan pembahasan mengenai limit tingkat SBMPTN. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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No. 1

Diketahui $f(x)=x^2+ax+b$ dengan $f(3)=1$. Jika $\displaystyle\lim_{x\to3}\dfrac{x-3}{f(x)-f(3)}=\dfrac12$ maka $a+b=$ ....

1. $8$
2. $0$
   
3. $-2$
   

4. $-4$
5. $-8$
   

$f'(x)=2x+a$

$\begin{aligned} \displaystyle\lim_{x\to3}\dfrac{x-3}{f(x)-f(3)}&=\dfrac12\\ \displaystyle\lim_{x\to3}\dfrac1{f'(x)}&=\dfrac12\\ \dfrac1{f'(3)}&=\dfrac12\\ \dfrac1{2(3)+a}&=\dfrac12\\ \dfrac1{6+a}&=\dfrac12\\ 6+a&=2\\ a&=-4 \end{aligned}$

$\begin{aligned} f(x)&=x^2+(-4)x+b\\ &=x^2-4x+b \end{aligned}$

$\begin{aligned} f(3)&=1\\ 3^2-4(3)+b&=1\\ 9-12+b&=1\\ -3+b&=1\\ b&=4 \end{aligned}$

$\begin{aligned} a+b&=-4+4\\ &=0 \end{aligned}$

Jadi, $a+b=0$.
JAWAB: B

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No. 2

Diketahui suku banyak ${f(x)=ax^2-(a+b)x-3}$ habis dibagi ${x+1}$. Jika $\displaystyle\lim_{x\to-1}\dfrac{f(x)}{x^2-x-2}=2$, maka nilai $a+b$ adalah ....

1. $-1$
2. $2$
   
3. $5$
   

4. $4$
5. $0$
   

habis dibagi $x+1$ artinya $f(-1)=0$

$\begin{aligned} f(-1)&=0\\ a(-1)^2-(a+b)(-1)-3&=0\\ a+a+b-3&=0\\ 2a+b&=3 \end{aligned}$

$f'(x)=2ax-a-b$

$\begin{aligned} \displaystyle\lim_{x\to-1}\dfrac{f(x)}{x^2-x-2}&=2\\ \displaystyle\lim_{x\to-1}\dfrac{f'(x)}{2x-1}&=2\\ \dfrac{f'(-1)}{2(-1)-1}&=2\\ \dfrac{2a(-1)-a-b}{-3}&=2\\ -2a-a-b&=-6\\ 3a+b&=6 \end{aligned}$

$\begin{aligned} 3a+b&=6\\ 2a+b&=3\qquad-\\\hline a&=3 \end{aligned}$

$\begin{aligned} 2a+b&=3\\ 2(3)+b&=3\\ b&=-3 \end{aligned}$

$\begin{aligned} a+b&=3+(-3)\\ &=0 \end{aligned}$

Jadi, nilai $a+b$ adalah 0.

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No. 3

$\displaystyle\lim_{x\to1}\left(\dfrac1{1-x}-\dfrac2{x-x^3}\right)$ =

2. 
   
3. 
   

5. 
   

UM UGM '05 Kode 621

$\begin{aligned} \displaystyle\lim_{x\to1}\left(\dfrac1{1-x}-\dfrac2{x-x^3}\right)&=\displaystyle\lim_{x\to1}\left(\dfrac1{1-x}-\dfrac2{x(1+x)(1-x)}\right)\\[8pt] &=\displaystyle\lim_{x\to1}\left(\dfrac{x(1+x)}{x(1+x)(1-x)}-\dfrac2{x(1+x)(1-x)}\right)\\[8pt] &=\displaystyle\lim_{x\to1}\dfrac{x+x^2-2}{x(1+x)(1-x)}\\[8pt] &=\displaystyle\lim_{x\to1}\dfrac{(x+2)(x-1)}{x(1+x)(1-x)}\\[8pt] &=\displaystyle\lim_{x\to1}\dfrac{-(x+2)(1-x)}{x(1+x)(1-x)}\\[8pt] &=\dfrac{-(1+2)}{1(1+1)}\\[8pt] &=\dfrac{-3}2\\ &=\boxed{\boxed{-\dfrac32}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to1}\left(\dfrac1{1-x}-\dfrac2{x-x^3}\right)=-\dfrac32$.
JAWAB: A

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No. 12

$\displaystyle\lim_{x\to5}\dfrac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}=$

1. $\sqrt3+\sqrt2$
2. $5-2\sqrt6$
   
3. $2\sqrt6$
   

4. $5$
5. $5+2\sqrt6$
   

GRUP WHATSAPP

$\begin{aligned} \displaystyle\lim_{x\to5}\dfrac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}&=\displaystyle\lim_{x\to5}\dfrac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}\cdot\dfrac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x+2\sqrt{x+1}}}\\[8pt] &=\displaystyle\lim_{x\to5}\dfrac{x+2\sqrt{x+1}}{\sqrt{x^2-4(x+1)}}\\[8pt] &=\dfrac{5+2\sqrt{5+1}}{\sqrt{5^2-4(5+1)}}\\[8pt] &=\dfrac{5+2\sqrt6}{\sqrt{25-24}}\\[8pt] &=\dfrac{5+2\sqrt6}{\sqrt1}\\ &=\boxed{\boxed{5+2\sqrt6}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to5}\dfrac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}$.
JAWAB: E

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13

Jika $f(x)=ax+b$ dan $\displaystyle\lim_{x\to4}\left(\dfrac{3x\cdot f(x)}{x-4}\right)=24$, maka nilai $f(5)=$

1. $0$
2. $1$
   
3. $2$
   

4. $3$
5. $4$
   

CARA 1

$\begin{aligned} 3(4)\cdot f(4)&=0\\ 12(4a+b)&=0\\ 4a+b&=0 \end{aligned}$

Gunakan aturan L'hopital
$\begin{aligned} \displaystyle\lim_{x\to4}\left(\dfrac{3x\cdot f(x)}{x-4}\right)&=24\\[8pt] \displaystyle\lim_{x\to4}\left(\dfrac{3f(x)+3x\cdot f'(x)}1\right)&=24\\[8pt] 3f(4)+3(4)f'(4)&=24\\ 3(4a+b)+12(a)&=24\\ 3(0)+12a&=24\\ 12a&=24\\ a&=2 \end{aligned}$

$\begin{aligned} 4a+b&=0\\ 4(2)+b&=0\\ 8+b&=0\\ b&=-8 \end{aligned}$

$f(x)=2x-8$

$\begin{aligned} f(5)&=2(5)-8\\ &=10-8\\ &=\boxed{\boxed{2}} \end{aligned}$

CARA 2

$f(x)$ adalah fungsi linier, dan $x-4$ harus menjadi salah satu faktornya, sehingga bisa kita tulis $f(x)=p(x-4)$
$\begin{aligned} \displaystyle\lim_{x\to4}\left(\dfrac{3x\cdot f(x)}{x-4}\right)&=24\\[8pt] \displaystyle\lim_{x\to4}\left(\dfrac{3x\cdot p(x-4)}{x-4}\right)&=24\\[8pt] \displaystyle\lim_{x\to4}3xp&=24\\ 3(4)p&=24\\ 12p&=24\\ p&=2 \end{aligned}$

$f(x)=2(x-4)$

$\begin{aligned} f(5)&=2(5-4)\\ &=2(1)\\ &=\boxed{\boxed{2}} \end{aligned}$

Jadi, $f(5)=2$.
JAWAB: C

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No. 6

Nilai $\displaystyle\lim_{x\to a}\left(5f(x)-3g(x)\right)=11$ dan $\displaystyle\lim_{x\to a}\left(2f(x)+3g(x)\right)=17$, maka nilai $\displaystyle\lim_{x\to a}\left(f(x)\cdot g(x)\right)=$

1. $21$
2. $16$
   
3. $15$
   

4. $14$
5. $12$
   

$\begin{aligned} \displaystyle\lim_{x\to a}\left(5f(x)-3g(x)\right)&=11\\ \displaystyle\lim_{x\to a}\left(2f(x)+3g(x)\right)&=17\qquad&\color{red}{+}\\\hline \displaystyle\lim_{x\to a}7f(x)&=28\\ \displaystyle\lim_{x\to a}f(x)&=4 \end{aligned}$

$\begin{aligned} \displaystyle\lim_{x\to a}\left(5f(x)-3g(x)\right)&=11\\ 5(4)-\displaystyle\lim_{x\to a}3g(x)&=11\\ 20-\displaystyle\lim_{x\to a}3g(x)&=11\\ -\displaystyle\lim_{x\to a}3g(x)&=-9\\ \displaystyle\lim_{x\to a}3g(x)&=9\\ \displaystyle\lim_{x\to a}g(x)&=3 \end{aligned}$

$\begin{aligned} \displaystyle\lim_{x\to a}\left(f(x)\cdot g(x)\right)&=4\cdot3\\ &=\boxed{\boxed{12}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to a}\left(f(x)\cdot g(x)\right)=12$.
JAWAB: E

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No. 7

Jika $\displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2}{x-1}=M$ maka $\displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2x}{x^2+2x-3}=$ ....

1. $2M-1$
2. $\dfrac12(M-1)$
3. $\dfrac14(M-2)$

4. $2M+2$
5. $4M-2$

CARA 1

$\begin{aligned} \displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2}{x-1}&=M\\[8pt] \displaystyle\lim_{x\to1}\dfrac{\dfrac{4ax^3}{2\sqrt{ax^4+b}}}1&=M\\[8pt] \displaystyle\lim_{x\to1}\dfrac{4ax^3}{2\sqrt{ax^4+b}}&=M\qquad\color{red}{\text{L'hopital}} \end{aligned}$

$\begin{aligned} \displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2x}{x^2+2x-3}&=\displaystyle\lim_{x\to1}\dfrac{\dfrac{4ax^3}{2\sqrt{ax^4+b}}-2}{2x+2}\qquad\color{red}{\text{L'hopital}}\\[8pt] &=\dfrac{M-2}{2(1)+2}\\[8pt] &=\dfrac{M-2}4\\ &=\boxed{\boxed{\dfrac14(M-2)}} \end{aligned}$

CARA 2

$\begin{aligned} \displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2x}{x^2+2x-3}&=\displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2-2x+2}{(x-1)(x+3)}\\[8pt] &=\displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2-2(x-1)}{(x-1)(x+3)}\\[8pt] &=\displaystyle\lim_{x\to1}\dfrac{\dfrac{\sqrt{ax^4+b}-2}{x-1}-2}{x+3}\\[8pt] &=\dfrac{M-2}{1+3}\\[8pt] &=\dfrac{M-2}4\\ &=\boxed{\boxed{\dfrac14(M-2)}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2x}{x^2+2x-3}=\dfrac14(M-2)$.
JAWAB: C

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No. 8

Diketahui suku banyak $g(x)=ax^2+(a-b)x+1$ habis dibagi $x+1$. Jika $\displaystyle\lim_{x\to-1}\dfrac{g(x)}{x^2+3x+2}=4$, maka nilai $2a-b$ adalah

1. $3$
2. $4$
   
3. $-5$
   

4. $-6$
5. $-7$
   

$g(x)=ax^2+(a-b)x+1$ habis dibagi $x+1$ berarti
$\begin{aligned} g(-1)&=0\\ a(-1)^2+(a-b)(-1)+1&=0\\ a-a+b+1&=0\\ b&=-1 \end{aligned}$

$\begin{aligned} g(x)&=ax^2+(a-b)x+1\\ &=ax^2+(a-(-1))x+1\\ &=ax^2+(a+1)x+1 \end{aligned}$

$\begin{aligned} \displaystyle\lim_{x\to-1}\dfrac{ax^2+(a+1)x+1}{x^2+3x+2}&=4\\[8pt] \displaystyle\lim_{x\to-1}\dfrac{2ax+a+1}{2x+3}&=4\\[8pt] \dfrac{2a(-1)+a+1}{2(-1)+3}&=4\\[8pt] \dfrac{-2a+a+1}{1}&=4\\[8pt] -a+1&=4\\ a&=-3 \end{aligned}$

$\begin{aligned} 2a-b&=2(-3)-(-1)\\ &=-6+1\\ &=-5 \end{aligned}$

Jadi, $2a-b=-5$.
JAWAB: C

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No. 9

Nilai $\displaystyle\lim_{x\to a}\left(5f(x)-3g(x)\right)= 11$ dan $\displaystyle\lim_{x\to a}\left(2f(x)+3g(x)\right)= 17$, maka nilai $\displaystyle\lim_{x\to a}\left(4f(x)\right)=$

1. $16$
2. $18$
   
3. $20$
   

4. $22$
5. $24$
   

$\begin{aligned} \displaystyle\lim_{x\to a}\left(5f(x)-3g(x)\right)&= 11\\ 5\displaystyle\lim_{x\to a}f(x)-3\displaystyle\lim_{x\to a}g(x)&=11 \end{aligned}$

$\begin{aligned} \displaystyle\lim_{x\to a}\left(2f(x)+3g(x)\right)&= 17\\ 2\displaystyle\lim_{x\to a}f(x)+3\displaystyle\lim_{x\to a}g(x)&=17 \end{aligned}$

$\begin{aligned} 5\displaystyle\lim_{x\to a}f(x)-3\displaystyle\lim_{x\to a}g(x)&=11\\ 2\displaystyle\lim_{x\to a}f(x)+3\displaystyle\lim_{x\to a}g(x)&=17\qquad\color{red}{+}\\\hline 7\displaystyle\lim_{x\to a}f(x)&=28\\ \displaystyle\lim_{x\to a}f(x)&=4\\ \displaystyle\lim_{x\to a}\left(4f(x)\right)&=\boxed{\boxed{16}} \end{aligned}$

Jadi, $\displaystyle\lim_{x\to a}\left(4f(x)\right)=16$.
JAWAB: A

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No. 10

Diketahui ${\displaystyle\lim_{x\to3}\dfrac{f(x)\cdot g(x)-5g(x)+f(x)-5}{\left(f(x)-5\right)(x-3)}=0}$. Nilai $g'(3)$ adalah

1. $-1$
2. $1$
   
3. $-3$
   

4. $3$
5. $0$
   

$\begin{aligned} \displaystyle\lim_{x\to3}\dfrac{f(x)\cdot g(x)-5g(x)+f(x)-5}{\left(f(x)-5\right)(x-3)}&=0\\[8pt] \displaystyle\lim_{x\to3}\dfrac{\left(f(x)-5\right)\left(g(x)+1\right)}{\left(f(x)-5\right)(x-3)}&=0\\[8pt] \displaystyle\lim_{x\to3}\dfrac{g(x)+1}{x-3}&=0\\[8pt] \displaystyle\lim_{x\to3}\dfrac{g'(x)}1&=0&\qquad\color{red}{\text{l'hopital}}\\[8pt] \displaystyle\lim_{x\to3}g'(x)&=0\\ g'(3)&=0 \end{aligned}$

Jadi, $g'(3)=0$.
JAWAB: E

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