Soal UAS Kalkulus I Kelas B

7:24:00 PM
Kemarin Saya ditanya teman-teman atau adik-adik leting saya tentang jawaban soal UAS Kalkulus I. Saya mencoba lihat soalnya ternyata cukup mudah saja kok. Makanya saya bagi disini. Bagi anda yang ingin melihatnya let's go....


1.  Buktikan bahwa ${\displaystyle \lim_{x\to1}\frac{14x^{2}-20x+6}{x-1}=8}$


Jawab:
Pertama-tama kita mencari $\delta$ sedemikian hingga
\[
0<|x-1|<\delta\Rightarrow\bigg|\frac{14x^{2}-20x+6}{x-1}-8\bigg|<\varepsilon
\]Untuk $x\neq1$ maka
\begin{eqnarray*}
\bigg|\frac{14x^{2}-20x+6}{x-1}-8\bigg|<\varepsilon & \Longleftrightarrow & \bigg|\frac{(14x-6)(x-1)}{x-1}-8\bigg|<\varepsilon\\
& \Longleftrightarrow & \bigg|(14x-6)-8\bigg|<\varepsilon\\
& \Longleftrightarrow & |14x-14|<\varepsilon\\
& \Longleftrightarrow & |14(x-1)|<\varepsilon\\
& \Longleftrightarrow & |14||x-1|<\varepsilon\\
& \Longleftrightarrow & |x-1|<\frac{\varepsilon}{14}
\end{eqnarray*}Hal ini menunjukkan bahwa $\delta={\displaystyle \frac{\varepsilon}{14}}$ akan memenuhi.Andaikan $\varepsilon>0$ ada $\delta={\displaystyle \frac{\varepsilon}{14}}$ maka $0<|x-1|<\delta$ sedemikian hingga\begin{eqnarray*}
\bigg|\frac{14x^{2}-20x+6}{x-1}-8\bigg| & = & \bigg|\frac{(14x-6)(x-1)}{x-1}-8\bigg|\\
& = & |(14x-6)-8|\\
& = & |14x-14|\\
& = & |14(x-1)|\\
& = & 14|x-1|\\
& < & 14\delta\\
& = & 14\cdot\frac{\varepsilon}{14}=\varepsilon
\end{eqnarray*}Sehingga terbukti bahwa ${\displaystyle \lim_{x\to1}\frac{14x^{2}-20x+6}{x-1}=8}$


2.   Hitunglah limit-limit berikut (jika ada)
a. ${\displaystyle \lim_{x\to1}}{\displaystyle \frac{(3x+4)(2x-2)^{3}}{(x-1)^{2}}}$
b. ${\displaystyle \lim_{\theta\to0}}{\displaystyle \frac{\tan(5\theta)}{\sin(2\theta)}}$
c. ${\displaystyle \lim_{n\to\infty}}{\displaystyle \frac{n}{n^{2}+1}}$
d. ${\displaystyle \lim_{x\to3}}{\displaystyle \frac{3+x}{3-x}}$


Jawab:
a. ${\displaystyle \lim_{x\to1}}{\displaystyle \frac{(3x+4)(2x-2)^{3}}{(x-1)^{2}}}$
\begin{eqnarray*}
{\displaystyle \lim_{x\to1}}{\displaystyle \frac{(3x+4)(2x-2)^{3}}{(x-1)^{2}}} & = & {\displaystyle \lim_{x\to1}}{\displaystyle \frac{(3x+4)(2(x-1))^{3}}{(x-1)^{2}}}\\
& = & {\displaystyle \lim_{x\to1}}{\displaystyle \frac{(3x+4)8(x-1)^{3}}{(x-1)^{2}}}\\
& = & {\displaystyle \lim_{x\to1}}{\displaystyle 8(3x+4)(x-1)}\\
& = & 0
\end{eqnarray*}
b. ${\displaystyle \lim_{\theta\to0}}{\displaystyle \frac{\tan(5\theta)}{\sin(2\theta)}}$
\begin{eqnarray*}
{\displaystyle \lim_{\theta\to0}}{\displaystyle \frac{\tan(5\theta)}{\sin(2\theta)}} & = & {\displaystyle \lim_{\theta\to0}}{\displaystyle \frac{{\displaystyle \frac{\sin(5\theta)}{\cos(5\theta)}}}{\sin(2\theta)}}\\
& = & {\displaystyle \lim_{\theta\to0}}{\displaystyle \frac{\sin(5\theta)}{\cos(5\theta)\cdot\sin(2\theta)}}\\
& = & {\displaystyle \lim_{\theta\to0}}{\displaystyle \frac{1}{\cos(5\theta)}\cdot\frac{\sin(5\theta)}{\sin(2\theta)}}\\
& = & {\displaystyle \lim_{\theta\to0}}{\displaystyle \frac{1}{\cos(5\theta)}\cdot\sin(5\theta)\cdot\frac{1}{\sin(2\theta)}}\\
& = & {\displaystyle \lim_{\theta\to0}}{\displaystyle \frac{1}{\cos(5\theta)}\cdot5\cdot\frac{\sin(5\theta)}{5\theta}\cdot\frac{1}{2}\cdot\frac{2\theta}{\sin(2\theta)}}\\
& = & \frac{5}{2}\lim_{\theta\to0}\frac{1}{\cos(5\theta)}\cdot\frac{\sin(5\theta)}{5\theta}\cdot\frac{2\theta}{\sin(2\theta)}\\
& = & \frac{5}{2}\cdot1\cdot1\cdot1\\
& = & \frac{5}{2}
\end{eqnarray*}
c. ${\displaystyle \lim_{n\to\infty}}{\displaystyle \frac{n}{n^{2}+1}}$
\begin{eqnarray*}
{\displaystyle \lim_{n\to\infty}}{\displaystyle \frac{n}{n^{2}+1}} & = & {\displaystyle \lim_{n\to\infty}}{\displaystyle \frac{{\displaystyle \frac{n}{n^{2}}}}{{\displaystyle \frac{n^{2}}{n^{2}}}+{\displaystyle \frac{1}{n^{2}}}}}\\
& = & \lim_{n\to\infty}\frac{{\displaystyle \frac{1}{n}}}{{\displaystyle 1+\frac{1}{n^{2}}}}\\
& = & \frac{0}{1+0}\\
& = & 0
\end{eqnarray*}
d. ${\displaystyle \lim_{x\to3}}{\displaystyle \frac{3+x}{3-x}}$.Limit tersebut tidak ada. Dengan menggunakan pendekatan limit dibeberapa titik maka kita dapatkan


Terlihat bahwa limit kiri tidak sama dengan limit kanan sehingga $\lim_{x\to3}f(x)$ tidak ada.


3.  Andaikan $f(x)=\begin{cases}
-1 & \text{jika }x\leq0\\
ax+b & \text{jika }0<x<1\\
1 & \text{jika }x\geq1
\end{cases}$Tentukan $a$ dan $b$ sehingga $f$ kontinu dimana-mana.


Jawab:


Untuk $x=0$ maka kita dapatkan $a(0)+b=-1$ dan untuk $x=1$ maka kita dapatkan $a(1)+b=1$ Sehingga
\begin{eqnarray*}
a(0)+b & = & -1\\
0+b & = & -1\\
b & = & -1
\end{eqnarray*}Untuk $b=-1$ maka
\begin{eqnarray*}
a(1)+b & = & 1\\
a+(-1) & = & 1\\
a & = & 2
\end{eqnarray*}Sehingga kita dapatkan $a=2$ dan $b=-1$.


4.  Dengan menggunakan $f'(x)={\displaystyle \lim_{h\to0}\frac{\left[f(x+h)-f(x)\right]}{h}}$ carilah turunan dari $f(x)={\displaystyle \frac{2x-1}{x-4}}$


Jawab:
\begin{eqnarray*}
f(x) & = & {\displaystyle \frac{2x-1}{x-4}}\\
f'(x) & = & \lim_{h\to0}\frac{\left[f(x+h)-f(x)\right]}{h}\\
& = & \lim_{h\to0}\frac{\left[{\displaystyle \frac{2(x+h)-1}{(x+h)-4}}-{\displaystyle \frac{2x-1}{x-4}}\right]}{h}\\
& = & \lim_{h\to0}\frac{\left[{\displaystyle \frac{\left(2x+2h-1\right)\left(x-4\right)-\left(2x-1\right)\left(x+h-4\right)}{\left(x+h-4\right)\left(x-4\right)}}\right]}{h}\\
& = & \lim_{h\to0}\left[{\displaystyle \frac{\left(2x+2h-1\right)\left(x-4\right)-\left(2x-1\right)\left(x+h-4\right)}{\left(x+h-4\right)\left(x-4\right)}}\right]\cdot\frac{1}{h}\\
& = & \lim_{h\to0}\left[{\displaystyle \frac{\left(2x^{2}-8x+2xh-8h-x+4\right)-\left(2x^{2}+2xh-8x-x-h+4\right)}{\left(x+h-4\right)\left(x-4\right)}}\right]\cdot\frac{1}{h}\\
& = & \lim_{h\to0}\left[{\displaystyle \frac{\left(2x^{2}-9x+2xh-8h+4\right)-\left(2x^{2}+2xh-9x-h+4\right)}{\left(x+h-4\right)\left(x-4\right)}}\cdot\frac{1}{h}\right]\\
& = & \lim_{h\to0}\left[{\displaystyle \frac{\left(2x^{2}-9x+2xh-8h+4\right)-\left(2x^{2}+2xh-9x-h+4\right)}{\left(x+h-4\right)\left(x-4\right)}}\cdot\frac{1}{h}\right]\\
& = & \lim_{h\to0}\left[\frac{-7h}{\left(x+h-4\right)\left(x-4\right)}\cdot\frac{1}{h}\right]\\
& = & \lim_{h\to0}\left[\frac{-7}{\left(x+h-4\right)\left(x-4\right)}\right]\\
& = & \left[\frac{-7}{\left(x-4\right)\left(x-4\right)}\right]\\
& = & -\frac{7}{(x-4)^{2}}
\end{eqnarray*}


5. Tentukan turunan dari fungsi-fungsi berikut.
a. $y=\sin^{4}(x^{2}+3x)$
b. $xy^{2}+yx^{2}=1$


Jawab:


a. $y=\sin^{4}(x^{2}+3x)$
\begin{eqnarray*}
y & = & \sin^{4}(x^{2}+3x)\\
y' & = & 4\sin^{3}(x^{2}+3x)\cos(x^{2}+3x)(2x+3)\\
y' & = & (8x+12)\sin^{3}(x^{2}+3x)\cos(x^{2}+3x)
\end{eqnarray*}
b. $xy^{2}+yx^{2}=1$
\begin{eqnarray*}
xy^{2}+yx^{2} & = & 1\\
y^{2}dx+2xydy+x^{2}dy+2xydx & = & 0\\
dx(y^{2}+2xy)+dy(2xy+x^{2}) & = & 0\\
dy(2xy+x^{2}) & = & -dx(y^{2}+2xy)\\
\frac{dy}{dx} & = & -\frac{(y^{2}+2xy)}{(2xy+x^{2})}
\end{eqnarray*}
Jawaban diatas belum tentu benar lho. Makanya kalau ada yang salah langsung aja di kritik yah di kotak komentar di bawah. Terima kasih atas kunjungannya...Jika anda ingin download Versi PDF nya silahkan klik di sini

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