SBMPTN Zone : Limit [2]

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

Tipe:

Standar SBMPTN HOTS

No.

{\displaystyle\lim_{x\to2}\dfrac{x^n-2^n}{x^{\frac{n}3}-2^{\frac{n}3}}=3\sqrt[3]{16}}, tentukan nilai n

GRUP TELEGRAM

CARA 1 : PEMFAKTORAN

$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x^n-2^n}{x^{\frac{n}{3}}-2^{\frac{n}{3}}}}}& =3\sqrt[3]{16}\\ {\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{{\left(x^{\frac{n}{3}}\right)}^3-{\left(2^{\frac{n}{3}}\right)}^3}{x^{\frac{n}{3}}-2^{\frac{n}{3}}}}}& =3\sqrt[3]{16}\\ {\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{(x^{\frac{n}{3}}-2^{\frac{n}{3}})(x^{\frac{2n}{3}}+x^{\frac{n}{3}}{2}^{\frac{n}{3}}+2^{\frac{2n}{3}})}{x^{\frac{n}{3}}-2^{\frac{n}{3}}}}}& =3\sqrt[3]{16}\\ {\displaystyle \underset{x\to 2}{lim}(x^{\frac{2n}{3}}+x^{\frac{n}{3}}{2}^{\frac{n}{3}}+2^{\frac{2n}{3}})}& =3\sqrt[3]{16}\\ 2^{\frac{2n}{3}}+2^{\frac{n}{3}}{2}^{\frac{n}{3}}+2^{\frac{2n}{3}}& =3\sqrt[3]{16}\\ 2^{\frac{2n}{3}}+2^{\frac{2n}{3}}+2^{\frac{2n}{3}}& =3\sqrt[3]{16}\\ 3\cdot 2^{\frac{2n}{3}}& =3\sqrt[3]{16}\\ 3\sqrt[3]{2^{2n}}& =3\sqrt[3]{16}\\ 2^{2n}& =16\\ 2n& =4\\ n& =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$$

CARA 2 : L'HOPITAL

$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{x^n-2^n}{x^{\frac{n}{3}}-2^{\frac{n}{3}}}}}& =3\sqrt[3]{16}\\ {\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{n{x}^{n-1}}{{\displaystyle \frac{n}{3}}{x}^{\frac{n}{3}-1}}}}& =3\sqrt[3]{16}\\ {\displaystyle \underset{x\to 2}{lim}3x^{n-\frac{n}{3}}}& =3\sqrt[3]{16}\\ {\displaystyle \underset{x\to 2}{lim}3x^{\frac{2n}{3}}}& =3\sqrt[3]{16}\\ {\displaystyle \underset{x\to 2}{lim}3\sqrt[3]{x^{2n}}}& =3\sqrt[3]{16}\\ 3\sqrt[3]{2^{2n}}& =3\sqrt[3]{16}\\ 2^{2n}& =16\\ 2n& =4\\ n& =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$$

Jadi, n=2.

No.

Diketahui f(x)=\sqrt{x+3}. Nilai \displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-3h^2\right)}{h^2} adalah

1. \dfrac52
2. 5\sqrt6
3. \dfrac5{12}\sqrt6

4. \dfrac52\sqrt6
5. 5

SKMP September 2020

CARA BIASA

$\begin{array}{rl}{\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(3+2h^2)-f(3-3h^2)}{h^2}}}& ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{3+2h^2+3}-\sqrt{3-3h^2+3}}{h^2}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{\sqrt{6+2h^2}-\sqrt{6-3h^2}}{h^2}}\cdot {\displaystyle \frac{\sqrt{6+2h^2}+\sqrt{6-3h^2}}{\sqrt{6+2h^2}+\sqrt{6-3h^2}}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{(6+2h^2)-(6-3h^2)}{h^2(\sqrt{6+2h^2}+\sqrt{6-3h^2})}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{6+2h^2-6+3h^2}{h^2(\sqrt{6+2h^2}+\sqrt{6-3h^2})}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{5h^2}{h^2(\sqrt{6+2h^2}+\sqrt{6-3h^2})}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{5}{\sqrt{6+2h^2}+\sqrt{6-3h^2}}}}\\ & ={\displaystyle \frac{5}{\sqrt{6+2(0{)}^2}+\sqrt{6-3(0{)}^2}}}\\ & ={\displaystyle \frac{5}{\sqrt{6}+\sqrt{6}}}\\ & ={\displaystyle \frac{5}{2\sqrt{6}}}\cdot {\displaystyle \frac{\sqrt{6}}{\sqrt{6}}}\\ & ={\displaystyle \frac{5}{2(6)}}\sqrt{6}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{5}{12}}\sqrt{6}}}}}\end{array}$

CARA CEPAT

f'(x)=\dfrac1{2\sqrt{x+3}}

$\begin{array}{rl}{\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{f(3+2h^2)-f(3-3h^2)}{h^2}}}& ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{4h{f}^{\prime }(3+2h^2)-(-6h)f^{\prime }(3-3h^2)}{2h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}(2f^{\prime }(3+2h^2)+3f^{\prime }(3-3h^2))}\\ & =2f^{\prime }\left(3\right)+3f^{\prime }\left(3\right)\\ & =5f^{\prime }(3)\\ & =5\cdot {\displaystyle \frac{1}{2\sqrt{3+3}}}\\ & ={\displaystyle \frac{5}{2\sqrt{6}}}\cdot {\displaystyle \frac{\sqrt{6}}{\sqrt{6}}}\\ & ={\displaystyle \frac{5}{2(6)}}\sqrt{6}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{5}{12}}\sqrt{6}}}}}\end{array}$

Jadi, \displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-3h^2\right)}{h^2}=\dfrac5{12}\sqrt6. JAWAB: C

No.

Diketahui f(x)=ax+b dengan f^{-1}(11)=2 dan f^{-1}(8)=1 dengan f^{-1} menyatakan fungsi invers f.
Nilai \displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h= ....

1. 4
2. 5
3. 6

4. 7
5. 8

Ganesha Operation

$$\begin{array}{rl}{f}^{-1}(11)& =2\\ f(2)& =11\\ 2a+b& =11\end{array}$$

$$\begin{array}{rl}{f}^{-1}(8)& =1\\ f(1)& =8\\ a+b& =8\end{array}$$

$$\begin{array}{rl}2a+b& =11\\ a+b& =8-\\ a& =3\end{array}$$

$$\begin{array}{rl}a+b& =8\\ 3+b& =8\\ b& =5\end{array}$$

f(x)=3x+5

$$\begin{array}{rl}{\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{(3+h)f(3)-3f(3+h)}{h}}}& ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{(3+h)(3(3)+5)-3(3(3+h)+5)}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{(3+h)(9+5)-3(9+3h+5)}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{(3+h)(14)-3(14+3h)}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{42+14h-42+9h}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}{\displaystyle \frac{5h}{h}}}\\ & ={\displaystyle \underset{h\to 0}{lim}5}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5}}}}\end{array}$$

Jadi, \displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h=5.

No.

Jika nilai \displaystyle\lim_{x\to1}\dfrac{\sqrt{ax+b}-3}{x-1}=\dfrac23, maka nilai {8a + 2b} adalah

1. 42
2. 44
3. 46

4. 48
5. 50

SKMP Juli 2020

Karena \displaystyle\lim_{x\to1}(x-1)=0, maka
$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}(\sqrt{ax+b}-3)}& =0\\ {\displaystyle \underset{x\to 1}{lim}\sqrt{ax+b}}& =3\end{array}$

$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{\sqrt{ax+b}-3}{x-1}}}& ={\displaystyle \frac{2}{3}}\\ {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{d(\sqrt{ax+b}-3)}{d(x-1)}}}& ={\displaystyle \frac{2}{3}}\\ {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{{\displaystyle \frac{a}{2\sqrt{ax+b}}}}{1}}}& ={\displaystyle \frac{2}{3}}\\ {\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{a}{2\sqrt{ax+b}}}}& ={\displaystyle \frac{2}{3}}\\ {\displaystyle \frac{a}{2(3)}}& ={\displaystyle \frac{2}{3}}\\ {\displaystyle \frac{a}{6}}& ={\displaystyle \frac{2}{3}}\\ a& =6\cdot {\displaystyle \frac{2}{3}}\\ & =4\end{array}$

$\begin{array}{rl}{\displaystyle \underset{x\to 1}{lim}\sqrt{ax+b}}& =3\\ \sqrt{a(1)+b}& =3\\ a+b& =9\\ 2a+2b& =18\\ 6a+2a+2b& =6(4)+18\\ 8a+2b& =24+18\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 42}}}}\end{array}$

Jadi, {8a + 2b=42}. JAWAB: A

No.

Jika {\displaystyle\lim_{x\to2}\dfrac{\sqrt{ax^4+b}-1}{x-2}=3} maka nilai dari {\displaystyle\lim_{x\to2}\dfrac{2\sqrt{ax^4+b}-x}{x^2+2x-8}=}

1. 1
2. 2
3. \dfrac16

4. \dfrac56
5. \dfrac12

SKMP Februari 2021

Misal f(x)=\sqrt{ax^4+b}

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

$$\begin{array}{rl}{\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{2\sqrt{a{x}^4+b}-x}{x^2+2x-8}}}& ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{2f(x)-x}{x^2+2x-8}}}\\ & ={\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{2f^{\prime }(x)-1}{2x+2}}}\\ & ={\displaystyle \frac{2(3)-1}{2(2)+2}}\\ & ={\displaystyle \frac{6-1}{4+2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{5}{6}}}}}}\end{array}$$

Jadi, {\displaystyle\lim_{x\to2}\dfrac{2\sqrt{ax^4+b}-x}{x^2+2x-8}=\dfrac56}. JAWAB: D