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

$$\begin{array}{rl}{\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{x-3}{f(x)-f(3)}}}& ={\displaystyle \frac{1}{2}}\\ {\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{1}{f^{\prime }(x)}}}& ={\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{f^{\prime }(3)}}& ={\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{2(3)+a}}& ={\displaystyle \frac{1}{2}}\\ {\displaystyle \frac{1}{6+a}}& ={\displaystyle \frac{1}{2}}\\ 6+a& =2\\ a& =-4\end{array}$$

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

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

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

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

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

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

$$\begin{array}{rl}{\displaystyle \underset{x\to -1}{lim}{\displaystyle \frac{f(x)}{x^2-x-2}}}& =2\\ {\displaystyle \underset{x\to -1}{lim}{\displaystyle \frac{f^{\prime }(x)}{2x-1}}}& =2\\ {\displaystyle \frac{f^{\prime }(-1)}{2(-1)-1}}& =2\\ {\displaystyle \frac{2a(-1)-a-b}{-3}}& =2\\ -2a-a-b& =-6\\ 3a+b& =6\end{array}$$

$$\begin{array}{rl}3a+b& =6\\ 2a+b& =3-\\ a& =3\end{array}$$

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

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

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

$$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}({\displaystyle \frac{1}{1-x}}-{\displaystyle \frac{2}{x-x^3}})}& ={\displaystyle \underset{x\to 1}{lim}({\displaystyle \frac{1}{1-x}}-{\displaystyle \frac{2}{x(1+x)(1-x)}})}\\ & ={\displaystyle \underset{x\to 1}{lim}({\displaystyle \frac{x(1+x)}{x(1+x)(1-x)}}-{\displaystyle \frac{2}{x(1+x)(1-x)}})}\\ & ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{x+x^2-2}{x(1+x)(1-x)}}}\\ & ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{(x+2)(x-1)}{x(1+x)(1-x)}}}\\ & ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{-(x+2)(1-x)}{x(1+x)(1-x)}}}\\ & ={\displaystyle \frac{-(1+2)}{1(1+1)}}\\ & ={\displaystyle \frac{-3}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{3}{2}}}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to 5}{lim}{\displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}}}& ={\displaystyle \underset{x\to 5}{lim}{\displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x-2\sqrt{x+1}}}}\cdot {\displaystyle \frac{\sqrt{x+2\sqrt{x+1}}}{\sqrt{x+2\sqrt{x+1}}}}}\\ & ={\displaystyle \underset{x\to 5}{lim}{\displaystyle \frac{x+2\sqrt{x+1}}{\sqrt{x^2-4(x+1)}}}}\\ & ={\displaystyle \frac{5+2\sqrt{5+1}}{\sqrt{5^2-4(5+1)}}}\\ & ={\displaystyle \frac{5+2\sqrt{6}}{\sqrt{25-24}}}\\ & ={\displaystyle \frac{5+2\sqrt{6}}{\sqrt{1}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5+2\sqrt{6}}}}}\end{array}$$

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

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

Gunakan aturan L'hopital
$$\begin{array}{rl}{\displaystyle \underset{x\to 4}{lim}\left({\displaystyle \frac{3x\cdot f(x)}{x-4}}\right)}& =24\\ {\displaystyle \underset{x\to 4}{lim}\left({\displaystyle \frac{3f(x)+3x\cdot f^{\prime }(x)}{1}}\right)}& =24\\ 3f(4)+3(4)f^{\prime }(4)& =24\\ 3(4a+b)+12(a)& =24\\ 3(0)+12a& =24\\ 12a& =24\\ a& =2\end{array}$$

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

f(x)=2x-8

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

CARA 2

f(x) adalah fungsi linier, dan x-4 harus menjadi salah satu faktornya, sehingga bisa kita tulis f(x)=p(x-4)
$$\begin{array}{rl}{\displaystyle \underset{x\to 4}{lim}\left({\displaystyle \frac{3x\cdot f(x)}{x-4}}\right)}& =24\\ {\displaystyle \underset{x\to 4}{lim}\left({\displaystyle \frac{3x\cdot p(x-4)}{x-4}}\right)}& =24\\ {\displaystyle \underset{x\to 4}{lim}3xp}& =24\\ 3(4)p& =24\\ 12p& =24\\ p& =2\end{array}$$

f(x)=2(x-4)

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

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
   

$$\begin{array}{rl}{\displaystyle \underset{x\to a}{lim}(5f(x)-3g(x))}& =11\\ {\displaystyle \underset{x\to a}{lim}(2f(x)+3g(x))}& =17& +\\ {\displaystyle \underset{x\to a}{lim}7f(x)}& =28\\ {\displaystyle \underset{x\to a}{lim}f(x)}& =4\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{x\to a}{lim}(5f(x)-3g(x))}& =11\\ 5(4)-{\displaystyle \underset{x\to a}{lim}3g(x)}& =11\\ 20-{\displaystyle \underset{x\to a}{lim}3g(x)}& =11\\ -{\displaystyle \underset{x\to a}{lim}3g(x)}& =-9\\ {\displaystyle \underset{x\to a}{lim}3g(x)}& =9\\ {\displaystyle \underset{x\to a}{lim}g(x)}& =3\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{x\to a}{lim}(f(x)\cdot g(x))}& =4\cdot 3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 12}}}}\end{array}$$

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

$$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{a{x}^4+b}-2}{x-1}}}& =M\\ {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{{\displaystyle \frac{4a{x}^3}{2\sqrt{a{x}^4+b}}}}{1}}}& =M\\ {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{4a{x}^3}{2\sqrt{a{x}^4+b}}}}& =M\text{L'hopital}\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{a{x}^4+b}-2x}{x^2+2x-3}}}& ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{{\displaystyle \frac{4a{x}^3}{2\sqrt{a{x}^4+b}}}-2}{2x+2}}\text{L'hopital}}\\ & ={\displaystyle \frac{M-2}{2(1)+2}}\\ & ={\displaystyle \frac{M-2}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{4}}(M-2)}}}}\end{array}$$

CARA 2

$$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{a{x}^4+b}-2x}{x^2+2x-3}}}& ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{a{x}^4+b}-2-2x+2}{(x-1)(x+3)}}}\\ & ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{a{x}^4+b}-2-2(x-1)}{(x-1)(x+3)}}}\\ & ={\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{{\displaystyle \frac{\sqrt{a{x}^4+b}-2}{x-1}}-2}{x+3}}}\\ & ={\displaystyle \frac{M-2}{1+3}}\\ & ={\displaystyle \frac{M-2}{4}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{4}}(M-2)}}}}\end{array}$$

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
$$\begin{array}{rl}g(-1)& =0\\ a(-1{)}^2+(a-b)(-1)+1& =0\\ a-a+b+1& =0\\ b& =-1\end{array}$$


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

$$\begin{array}{rl}{\displaystyle \underset{x\to -1}{lim}{\displaystyle \frac{a{x}^2+(a+1)x+1}{x^2+3x+2}}}& =4\\ {\displaystyle \underset{x\to -1}{lim}{\displaystyle \frac{2ax+a+1}{2x+3}}}& =4\\ {\displaystyle \frac{2a(-1)+a+1}{2(-1)+3}}& =4\\ {\displaystyle \frac{-2a+a+1}{1}}& =4\\ -a+1& =4\\ a& =-3\end{array}$$

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

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
   

$$\begin{array}{rl}{\displaystyle \underset{x\to a}{lim}(5f(x)-3g(x))}& =11\\ 5{\displaystyle \underset{x\to a}{lim}f(x)-3{\displaystyle \underset{x\to a}{lim}g(x)}}& =11\end{array}$$

$$\begin{array}{rl}{\displaystyle \underset{x\to a}{lim}(2f(x)+3g(x))}& =17\\ 2{\displaystyle \underset{x\to a}{lim}f(x)+3{\displaystyle \underset{x\to a}{lim}g(x)}}& =17\end{array}$$

$$\begin{array}{rl}5{\displaystyle \underset{x\to a}{lim}f(x)-3{\displaystyle \underset{x\to a}{lim}g(x)}}& =11\\ 2{\displaystyle \underset{x\to a}{lim}f(x)+3{\displaystyle \underset{x\to a}{lim}g(x)}}& =17+\\ 7{\displaystyle \underset{x\to a}{lim}f(x)}& =28\\ {\displaystyle \underset{x\to a}{lim}f(x)}& =4\\ {\displaystyle \underset{x\to a}{lim}(4f(x))}& =\boxed{{\displaystyle \boxed{{\displaystyle 16}}}}\end{array}$$

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
   

$$\begin{array}{rl}{\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{f(x)\cdot g(x)-5g(x)+f(x)-5}{(f(x)-5)(x-3)}}}& =0\\ {\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{(f(x)-5)(g(x)+1)}{(f(x)-5)(x-3)}}}& =0\\ {\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{g(x)+1}{x-3}}}& =0\\ {\displaystyle \underset{x\to 3}{lim}{\displaystyle \frac{g^{\prime }(x)}{1}}}& =0& \text{l'hopital}\\ {\displaystyle \underset{x\to 3}{lim}{g}^{\prime }(x)}& =0\\ g^{\prime }(3)& =0\end{array}$$

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

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