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.

Tingkat Kesulitan:

Dasar SBMPTN Olimpiade

• 1
• 2

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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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