SNBT Zone : Limit Fungsi Aljabar [2]
Table of Contents
Tipe:
No.
\({\displaystyle\lim_{x\to2}\dfrac{x^n-2^n}{x^{\frac{n}3}-2^{\frac{n}3}}=3\sqrt[3]{16}}\), tentukan nilai nALTERNATIF PENYELESAIAN
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- \(\dfrac52\)
- \(5\sqrt6\)
- \(\dfrac5{12}\sqrt6\)
- \(\dfrac52\sqrt6\)
- 5
ALTERNATIF PENYELESAIAN
CARA BIASA
\(\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-3h^2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{\sqrt{3+2h^2+3}-\sqrt{3-3h^2+3}}{h^2}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{\sqrt{6+2h^2}-\sqrt{6-3h^2}}{h^2}\cdot{\color{red}\dfrac{\sqrt{6+2h^2}+\sqrt{6-3h^2}}{\sqrt{6+2h^2}+\sqrt{6-3h^2}}}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{\left(6+2h^2\right)-\left(6-3h^2\right)}{h^2\left(\sqrt{6+2h^2}+\sqrt{6-3h^2}\right)}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{6+2h^2-6+3h^2}{h^2\left(\sqrt{6+2h^2}+\sqrt{6-3h^2}\right)}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac{5h^2}{h^2\left(\sqrt{6+2h^2}+\sqrt{6-3h^2}\right)}\\[3.7pt] &=\displaystyle\lim_{h\to0}\dfrac5{\sqrt{6+2h^2}+\sqrt{6-3h^2}}\\[3.7pt] &=\dfrac5{\sqrt{6+2(0)^2}+\sqrt{6-3(0)^2}}\\[3.7pt] &=\dfrac5{\sqrt6+\sqrt6}\\[3.7pt] &=\dfrac5{2\sqrt6}\cdot{\color{red}\dfrac{\sqrt6}{\sqrt6}}\\[3.7pt] &=\dfrac5{2(6)}\sqrt6\\ &=\boxed{\boxed{\dfrac5{12}\sqrt6}} \end{aligned}\)CARA CEPAT
\(f'(x)=\dfrac1{2\sqrt{x+3}}\)\(\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{f\left(3+2h^2\right)-f\left(3-3h^2\right)}{h^2}&=\displaystyle\lim_{h\to0}\dfrac{4h\ f'\left(3+2h^2\right)-(-6h)\ f'\left(3-3h^2\right)}{2h}\\[3.7pt] &=\displaystyle\lim_{h\to0}\left(2\ f'\left(3+2h^2\right)+3\ f'\left(3-3h^2\right)\right)\\ &=2\ f'\left(3\right)+3\ f'\left(3\right)\\ &=5f'(3)\\ &=5\cdot\dfrac1{2\sqrt{3+3}}\\[3.7pt] &=\dfrac5{2\sqrt6}\cdot{\color{red}\dfrac{\sqrt6}{\sqrt6}}\\[3.7pt] &=\dfrac5{2(6)}\sqrt6\\ &=\boxed{\boxed{\dfrac5{12}\sqrt6}} \end{aligned}\)
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
JAWAB: C
No.
Diketahui- 4
- 5
- 6
- 7
- 8
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
f^{-1}(11)&=2\\
f(2)&=11\\
2a+b&=11
\end{aligned}
\begin{aligned}
f^{-1}(8)&=1\\
f(1)&=8\\
a+b&=8
\end{aligned}
\begin{aligned}
2a+b&=11\\
a+b&=8\qquad-\\\hline
a&=3
\end{aligned}
\begin{aligned}
a+b&=8\\
3+b&=8\\
b&=5
\end{aligned}\)
f(x) = 3x + 5
\(\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h&=\displaystyle\lim_{h\to0}\dfrac{(3+h)(3(3)+5)-3(3(3+h)+5)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{(3+h)(9+5)-3(9+3h+5)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{(3+h)(14)-3(14+3h)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{42+14h-42+9h}h\\ &=\displaystyle\lim_{h\to0}\dfrac{5h}h\\ &=\displaystyle\lim_{h\to0}5\\ &=\boxed{\boxed{5}} \end{aligned}\)
\(\begin{aligned} \displaystyle\lim_{h\to0}\dfrac{(3+h)f(3)-3f(3+h)}h&=\displaystyle\lim_{h\to0}\dfrac{(3+h)(3(3)+5)-3(3(3+h)+5)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{(3+h)(9+5)-3(9+3h+5)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{(3+h)(14)-3(14+3h)}h\\ &=\displaystyle\lim_{h\to0}\dfrac{42+14h-42+9h}h\\ &=\displaystyle\lim_{h\to0}\dfrac{5h}h\\ &=\displaystyle\lim_{h\to0}5\\ &=\boxed{\boxed{5}} \end{aligned}\)
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- 42
- 44
- 46
- 48
- 50
ALTERNATIF PENYELESAIAN
Karena \(\displaystyle\lim_{x\to1}(x-1)=0\), maka
\(\begin{aligned} \displaystyle\lim_{x\to1}\left(\sqrt{ax+b}-3\right)&=0\\ \displaystyle\lim_{x\to1}\sqrt{ax+b}&=3 \end{aligned}\)
\(\begin{aligned} \displaystyle\lim_{x\to1}\dfrac{\sqrt{ax+b}-3}{x-1}&=\dfrac23\\[3.7pt] \displaystyle\lim_{x\to1}\dfrac{d\left(\sqrt{ax+b}-3\right)}{d(x-1)}&=\dfrac23\\[3.7pt] \displaystyle\lim_{x\to1}\dfrac{\dfrac{a}{2\sqrt{ax+b}}}1&=\dfrac23\\[3.7pt] \displaystyle\lim_{x\to1}\dfrac{a}{2\sqrt{ax+b}}&=\dfrac23\\[3.7pt] \dfrac{a}{2(3)}&=\dfrac23\\[3.7pt] \dfrac{a}6&=\dfrac23\\[3.7pt] a&=6\cdot\dfrac23\\ &=4 \end{aligned}\)
\(\begin{aligned} \displaystyle\lim_{x\to1}\left(\sqrt{ax+b}-3\right)&=0\\ \displaystyle\lim_{x\to1}\sqrt{ax+b}&=3 \end{aligned}\)
\(\begin{aligned} \displaystyle\lim_{x\to1}\dfrac{\sqrt{ax+b}-3}{x-1}&=\dfrac23\\[3.7pt] \displaystyle\lim_{x\to1}\dfrac{d\left(\sqrt{ax+b}-3\right)}{d(x-1)}&=\dfrac23\\[3.7pt] \displaystyle\lim_{x\to1}\dfrac{\dfrac{a}{2\sqrt{ax+b}}}1&=\dfrac23\\[3.7pt] \displaystyle\lim_{x\to1}\dfrac{a}{2\sqrt{ax+b}}&=\dfrac23\\[3.7pt] \dfrac{a}{2(3)}&=\dfrac23\\[3.7pt] \dfrac{a}6&=\dfrac23\\[3.7pt] a&=6\cdot\dfrac23\\ &=4 \end{aligned}\)
\(\begin{aligned}
\displaystyle\lim_{x\to1}\sqrt{ax+b}&=3\\
\sqrt{a(1)+b}&=3\\
a+b&=9\\
2a+2b&=18\\
6a+2a+2b&=6(4)+18\\
8a+2b&=24+18\\
&=\boxed{\boxed{42}}
\end{aligned}\)
Jadi, 8a + 2b = 42 .
JAWAB: A
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
- 2
- \(\dfrac16\)
- \(\dfrac56\)
- \(\dfrac12\)
ALTERNATIF PENYELESAIAN
Misal \(f(x)=\sqrt{ax^4+b}\)
\(\begin{aligned} \displaystyle\lim_{x\to2}\dfrac{\sqrt{ax^4+b}-1}{x-2}&=3\\[3.7pt] \displaystyle\lim_{x\to2}\dfrac{f(x)-1}{x-2}&=3\\[3.7pt] \displaystyle\lim_{x\to2}\dfrac{f'(x)}1&=3\\[3.7pt] \displaystyle\lim_{x\to2}f'(x)&=3 \end{aligned}\)
\(\begin{aligned} \displaystyle\lim_{x\to2}\dfrac{\sqrt{ax^4+b}-1}{x-2}&=3\\[3.7pt] \displaystyle\lim_{x\to2}\dfrac{f(x)-1}{x-2}&=3\\[3.7pt] \displaystyle\lim_{x\to2}\dfrac{f'(x)}1&=3\\[3.7pt] \displaystyle\lim_{x\to2}f'(x)&=3 \end{aligned}\)
\(\begin{aligned}
\displaystyle\lim_{x\to2}\dfrac{2\sqrt{ax^4+b}-x}{x^2+2x-8}&=\displaystyle\lim_{x\to2}\dfrac{2f(x)-x}{x^2+2x-8}\\[3.7pt]
&=\displaystyle\lim_{x\to2}\dfrac{2f'(x)-1}{2x+2}\\[3.7pt]
&=\dfrac{2(3)-1}{2(2)+2}\\[3.7pt]
&=\dfrac{6-1}{4+2}\\
&=\boxed{\boxed{\dfrac56}}
\end{aligned}\)
Jadi, \({\displaystyle\lim_{x\to2}\dfrac{2\sqrt{ax^4+b}-x}{x^2+2x-8}=\dfrac56}\).
JAWAB: D
JAWAB: D
No.
Jika p dan q merupakan bilangan real sehingga $\displaystyle\lim_{x\to\frac13}\dfrac{\sqrt{9x^4+3}-\sqrt{\dfrac{p}{x^2}+q}}{3x-1}=L$, maka $\displaystyle\lim_{x\to3}\dfrac{\sqrt{3x^4+9}-\sqrt{px^6+qx^4}}{x^2-9}=$ ....- $-\dfrac32L$
- $-\dfrac{L}2$
- $\dfrac{L}2$
- $\dfrac32L$
- $\dfrac52L$
ALTERNATIF PENYELESAIAN
Misal $a=\dfrac1x$.
Jika $x\to3$, maka $a=\dfrac1x\to\dfrac13$
\(\begin{aligned} \displaystyle\lim_{x\to3}\dfrac{\sqrt{3x^4+9}-\sqrt{px^6+qx^4}}{x^2-9}&=\displaystyle\lim_{x\to3}\dfrac{\sqrt{x^4\left(3+\dfrac9{x^4}\right)}-\sqrt{x^4\left(px^2+q\right)}}{(x+3)(x-3)}\\[4pt] &=\displaystyle\lim_{x\to3}\dfrac{x^2\sqrt{3+\dfrac9{x^4}}-x^2\sqrt{px^2+q}}{-x(x+3)\left(\dfrac3x-1\right)}\\[4pt] &=\displaystyle\lim_{x\to3}\dfrac{x}{-(x+3)}\cdot\displaystyle\lim_{x\to3}\dfrac{\sqrt{\dfrac9{x^4}+3}-\sqrt{px^2+q}}{\dfrac3x-1}\\[4pt] &=\dfrac3{-(3+3)}\cdot\displaystyle\lim_{a\to\frac13}\dfrac{\sqrt{9a^4+3}-\sqrt{\dfrac{p}{a^2}+q}}{3a-1}\\[4pt] &=\dfrac3{-6}\cdot L\\ &=\color{blue}\boxed{\boxed{\color{black}-\dfrac{L}2}} \end{aligned}\)
Jika $x\to3$, maka $a=\dfrac1x\to\dfrac13$
\(\begin{aligned} \displaystyle\lim_{x\to3}\dfrac{\sqrt{3x^4+9}-\sqrt{px^6+qx^4}}{x^2-9}&=\displaystyle\lim_{x\to3}\dfrac{\sqrt{x^4\left(3+\dfrac9{x^4}\right)}-\sqrt{x^4\left(px^2+q\right)}}{(x+3)(x-3)}\\[4pt] &=\displaystyle\lim_{x\to3}\dfrac{x^2\sqrt{3+\dfrac9{x^4}}-x^2\sqrt{px^2+q}}{-x(x+3)\left(\dfrac3x-1\right)}\\[4pt] &=\displaystyle\lim_{x\to3}\dfrac{x}{-(x+3)}\cdot\displaystyle\lim_{x\to3}\dfrac{\sqrt{\dfrac9{x^4}+3}-\sqrt{px^2+q}}{\dfrac3x-1}\\[4pt] &=\dfrac3{-(3+3)}\cdot\displaystyle\lim_{a\to\frac13}\dfrac{\sqrt{9a^4+3}-\sqrt{\dfrac{p}{a^2}+q}}{3a-1}\\[4pt] &=\dfrac3{-6}\cdot L\\ &=\color{blue}\boxed{\boxed{\color{black}-\dfrac{L}2}} \end{aligned}\)
Jadi, $\displaystyle\lim_{x\to3}\dfrac{\sqrt{3x^4+9}-\sqrt{px^6+qx^4}}{x^2-9}=-\dfrac{L}2$.
JAWAB: B
JAWAB: B
Post a Comment