Soal UAS Kalkulus I Kelas B

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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*}


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