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SNBT Zone : Limit

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Idham Muqoddas
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Berikut ini adalah kumpulan soal mengenai Limit. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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StandarSNBTHOTS


No.

Diketahui f(x) = x2 + 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
  1. −4
  2. −8
SUMBER
ALTERNATIF PENYELESAIAN
f'(x) = 2x + a \begin{aligned} \displaystyle\lim_{x\to3}\dfrac{x-3}{f(x)-f(3)}&=\dfrac12\\[3.7pt] \displaystyle\lim_{x\to3}\dfrac1{f'(x)}&=\dfrac12\\[3.7pt] \dfrac1{f'(3)}&=\dfrac12\\[3.7pt] \dfrac1{2(3)+a}&=\dfrac12\\[3.7pt] \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\\ &=\boxed{\boxed{0}} \end{aligned}
Jadi, a + b = 0.
JAWAB: B

No.

Diketahui suku banyak f(x) = ax2 − (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
  1. 4
  2. 0
SUMBER
ALTERNATIF PENYELESAIAN
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\\[3.7pt] \displaystyle\lim_{x\to-1}\dfrac{f'(x)}{2x-1}&=2\\[3.7pt] \dfrac{f'(-1)}{2(-1)-1}&=2\\[3.7pt] \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.
JAWAB: E

No.

\displaystyle\lim_{x\to1}\left(\dfrac1{1-x}-\dfrac2{x-x^3}\right) =
  1. -\dfrac32
  2. -\dfrac23
  3. \dfrac23
  1. 1
  2. \dfrac32
UM UGM '05 Kode 621
ALTERNATIF PENYELESAIAN
\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)\\[3.7pt] &=\displaystyle\lim_{x\to1}\left(\dfrac{x(1+x)}{x(1+x)(1-x)}-\dfrac2{x(1+x)(1-x)}\right)\\[3.7pt] &=\displaystyle\lim_{x\to1}\dfrac{x+x^2-2}{x(1+x)(1-x)}\\[3.7pt] &=\displaystyle\lim_{x\to1}\dfrac{(x+2)(x-1)}{x(1+x)(1-x)}\\[3.7pt] &=\displaystyle\lim_{x\to1}\dfrac{-(x+2)(1-x)}{x(1+x)(1-x)}\\[3.7pt] &=\dfrac{-(1+2)}{1(1+1)}\\[3.7pt] &=\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

No.

\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
  1. 5
  2. 5+2\sqrt6
GRUP WHATSAPP
ALTERNATIF PENYELESAIAN
\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}}}\\[3.7pt] &=\displaystyle\lim_{x\to5}\dfrac{x+2\sqrt{x+1}}{\sqrt{x^2-4(x+1)}}\\[3.7pt] &=\dfrac{5+2\sqrt{5+1}}{\sqrt{5^2-4(5+1)}}\\[3.7pt] &=\dfrac{5+2\sqrt6}{\sqrt{25-24}}\\[3.7pt] &=\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

No.

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
  1. 3
  2. 4
SUMBER
ALTERNATIF PENYELESAIAN

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\\[3.7pt] \displaystyle\lim_{x\to4}\left(\dfrac{3f(x)+3x\cdot f'(x)}1\right)&=24\\[3.7pt] 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\\[3.7pt] \displaystyle\lim_{x\to4}\left(\dfrac{3x\cdot p(x-4)}{x-4}\right)&=24\\[3.7pt] \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

No.

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
  1. 14
  2. 12
SUMBER
ALTERNATIF PENYELESAIAN
\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&+\\\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

No.

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)
  1. 2M + 2
  2. 4M − 2
SUMBER
ALTERNATIF PENYELESAIAN

CARA 1

\begin{aligned} \displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2}{x-1}&=M\\[3.7pt] \displaystyle\lim_{x\to1}\dfrac{\dfrac{4ax^3}{2\sqrt{ax^4+b}}}1&=M\\[3.7pt] \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}}\\[3.7pt] &=\dfrac{M-2}{2(1)+2}\\[3.7pt] &=\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)}\\[3.7pt] &=\displaystyle\lim_{x\to1}\dfrac{\sqrt{ax^4+b}-2-2(x-1)}{(x-1)(x+3)}\\[3.7pt] &=\displaystyle\lim_{x\to1}\dfrac{\dfrac{\sqrt{ax^4+b}-2}{x-1}-2}{x+3}\\[3.7pt] &=\dfrac{M-2}{1+3}\\[3.7pt] &=\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

No.

Diketahui suku banyak g(x) = ax2 + (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
  1. −6
  2. −7
SUMBER
ALTERNATIF PENYELESAIAN
g(x) = ax2 + (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\\[3.7pt] \displaystyle\lim_{x\to-1}\dfrac{2ax+a+1}{2x+3}&=4\\[3.7pt] \dfrac{2a(-1)+a+1}{2(-1)+3}&=4\\[3.7pt] \dfrac{-2a+a+1}{1}&=4\\[3.7pt] -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

No.

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
  1. 22
  2. 24
SUMBER
ALTERNATIF PENYELESAIAN
\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&+\\\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

No.

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
  1. 3
  2. 0
SUMBER
ALTERNATIF PENYELESAIAN
\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\\[3.7pt] \displaystyle\lim_{x\to3}\dfrac{\left(f(x)-5\right)\left(g(x)+1\right)}{\left(f(x)-5\right)(x-3)}&=0\\[3.7pt] \displaystyle\lim_{x\to3}\dfrac{g(x)+1}{x-3}&=0\\[3.7pt] \displaystyle\lim_{x\to3}\dfrac{g'(x)}1&=0&\qquad\color{red}{\text{l'hopital}}\\[3.7pt] \displaystyle\lim_{x\to3}g'(x)&=0\\ g'(3)&=0 \end{aligned}
Jadi, g'(3) = 0.
JAWAB: E


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