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{aligned} \displaystyle\lim_{x\to2}\dfrac{x^n-2^n}{x^{\frac{n}3}-2^{\frac{n}3}}&=3\sqrt[3]{16}\\ \displaystyle\lim_{x\to2}\dfrac{\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\lim_{x\to2}\dfrac{\left(x^{\frac{n}3}-2^{\frac{n}3}\right)\left(x^{\frac{2n}3}+x^{\frac{n}3}2^{\frac{n}3}+2^{\frac{2n}3}\right)}{x^{\frac{n}3}-2^{\frac{n}3}}&=3\sqrt[3]{16}\\ \displaystyle\lim_{x\to2}\left(x^{\frac{2n}3}+x^{\frac{n}3}2^{\frac{n}3}+2^{\frac{2n}3}\right)&=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\cdot2^{\frac{2n}3}&=3\sqrt[3]{16}\\ 3\sqrt[3]{2^{2n}}&=3\sqrt[3]{16}\\ 2^{2n}&=16\\ 2n&=4\\ n&=\boxed{\boxed{2}} \end{aligned}

CARA 2 : L'HOPITAL

\begin{aligned} \displaystyle\lim_{x\to2}\dfrac{x^n-2^n}{x^{\frac{n}3}-2^{\frac{n}3}}&=3\sqrt[3]{16}\\ \displaystyle\lim_{x\to2}\dfrac{nx^{n-1}}{\dfrac{n}3x^{\frac{n}3-1}}&=3\sqrt[3]{16}\\ \displaystyle\lim_{x\to2}3x^{n-\frac{n}3}&=3\sqrt[3]{16}\\ \displaystyle\lim_{x\to2}3x^{\frac{2n}3}&=3\sqrt[3]{16}\\ \displaystyle\lim_{x\to2}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{\boxed{2}} \end{aligned}

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

\(\eqalign{ \displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-3h^2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{\sqrt{3+2h^2+3}-\sqrt{3-3h^2+3}}{h^2}\\ &=\displaystyle\lim_{h\to0}\dfrac{\sqrt{6+2h^2}-\sqrt{6-3h^2}}{h^2}\cdot{\color{red}\dfrac{\sqrt{6+2h^2}+\sqrt{6-3h^2}}{\sqrt{6+2h^2}+\sqrt{6-3h^2}}}\\ &=\displaystyle\lim_{h\to0}\dfrac{\left(6+2h^2\right)-\left(6-3h^2\right)}{h^2\left(\sqrt{6+2h^2}+\sqrt{6-3h^2}\right)}\\ &=\displaystyle\lim_{h\to0}\dfrac{6+2h^2-6+3h^2}{h^2\left(\sqrt{6+2h^2}+\sqrt{6-3h^2}\right)}\\ &=\displaystyle\lim_{h\to0}\dfrac{5h^2}{h^2\left(\sqrt{6+2h^2}+\sqrt{6-3h^2}\right)}\\ &=\displaystyle\lim_{h\to0}\dfrac5{\sqrt{6+2h^2}+\sqrt{6-3h^2}}\\ &=\dfrac5{\sqrt{6+2(0)^2}+\sqrt{6-3(0)^2}}\\ &=\dfrac5{\sqrt6+\sqrt6}\\ &=\dfrac5{2\sqrt6}\cdot{\color{red}\dfrac{\sqrt6}{\sqrt6}}\\ &=\dfrac5{2(6)}\sqrt6\\ &=\boxed{\boxed{\dfrac5{12}\sqrt6}} }\)

CARA CEPAT

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

\(\eqalign{ \displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-3h^2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{4h\ f'\left(3+2h^2\right)-(-6h)\ f'\left(3-3h^2\right)}{2h}\\ &=\displaystyle\lim_{h\to0}\left(2\ f'\left(3+2h^2\right)+3\ f'\left(3-3h^2\right)\right)\\ &=2\ f'\left(3\right)+3\ f'\left(3\right)\\ &=5f'(3)\\ &=5\cdot\dfrac1{2\sqrt{3+3}}\\ &=\dfrac5{2\sqrt6}\cdot{\color{red}\dfrac{\sqrt6}{\sqrt6}}\\ &=\dfrac5{2(6)}\sqrt6\\ &=\boxed{\boxed{\dfrac5{12}\sqrt6}} }\)

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{aligned} f^{-1}(11)&=2\\ f(2)&=11\\ 2a+b&=11 \end{aligned}

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

\begin{aligned} 2a+b&=11\\ a+b&=8\qquad-\\\hline a&=3 \end{aligned}

\begin{aligned} a+b&=8\\ 3+b&=8\\ b&=5 \end{aligned}

f(x)=3x+5

\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h&=\displaystyle\lim_{h\to0}\dfrac{(3+h)(3(3)+5)-3(3(3+h)+5)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{(3+h)(9+5)-3(9+3h+5)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{(3+h)(14)-3(14+3h)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{42+14h-42+9h}h\\ &=\displaystyle\lim_{h\to0}\dfrac{5h}h\\ &=\displaystyle\lim_{h\to0}5\\ &=\boxed{\boxed{5}} \end{aligned}

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
\(\eqalign{ \displaystyle\lim_{x\to1}\left(\sqrt{ax+b}-3\right)&=0\\ \displaystyle\lim_{x\to1}\sqrt{ax+b}&=3 }\)

\(\eqalign{ \displaystyle\lim_{x\to1}\dfrac{\sqrt{ax+b}-3}{x-1}&=\dfrac23\\ \displaystyle\lim_{x\to1}\dfrac{d\left(\sqrt{ax+b}-3\right)}{d(x-1)}&=\dfrac23\\ \displaystyle\lim_{x\to1}\dfrac{\dfrac{a}{2\sqrt{ax+b}}}1&=\dfrac23\\ \displaystyle\lim_{x\to1}\dfrac{a}{2\sqrt{ax+b}}&=\dfrac23\\ \dfrac{a}{2(3)}&=\dfrac23\\ \dfrac{a}6&=\dfrac23\\ a&=6\cdot\dfrac23\\ &=4 }\)

\(\eqalign{ \displaystyle\lim_{x\to1}\sqrt{ax+b}&=3\\ \sqrt{a(1)+b}&=3\\ a+b&=9\\ 2a+2b&=18\\ 6a+2a+2b&=6(4)+18\\ 8a+2b&=24+18\\ &=\boxed{\boxed{42}} }\)

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{aligned} \displaystyle\lim_{x\to2}\dfrac{\sqrt{ax^4+b}-1}{x-2}&=3\\ \displaystyle\lim_{x\to2}\dfrac{f(x)-1}{x-2}&=3\\ \displaystyle\lim_{x\to2}\dfrac{f'(x)}1&=3\\ \displaystyle\lim_{x\to2}f'(x)&=3 \end{aligned}

\begin{aligned} \displaystyle\lim_{x\to2}\dfrac{2\sqrt{ax^4+b}-x}{x^2+2x-8}&=\displaystyle\lim_{x\to2}\dfrac{2f(x)-x}{x^2+2x-8}\\ &=\displaystyle\lim_{x\to2}\dfrac{2f'(x)-1}{2x+2}\\ &=\dfrac{2(3)-1}{2(2)+2}\\ &=\dfrac{6-1}{4+2}\\ &=\boxed{\boxed{\dfrac56}} \end{aligned}

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