Soal UAS Kalkulus I Kelas B

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

Jawab:

Pertama-tama kita mencari $\delta$ sedemikian hingga $$0<|x-1|<\delta \Rightarrow |\frac{14x^2-20x+6}{x-1}-8|<\epsilon$$Untuk $x\ne 1$ maka $$\begin{array}{rcl}|\frac{14x^2-20x+6}{x-1}-8|<\epsilon & ⟺& |\frac{(14x-6)(x-1)}{x-1}-8|<\epsilon \\ & ⟺& |(14x-6)-8|<\epsilon \\ & ⟺& |14x-14|<\epsilon \\ & ⟺& |14(x-1)|<\epsilon \\ & ⟺& |14||x-1|<\epsilon \\ & ⟺& |x-1|<\frac{\epsilon }{14}\end{array}$$Hal ini menunjukkan bahwa $\delta ={\displaystyle \frac{\epsilon }{14}}$ akan memenuhi.Andaikan $\epsilon >0$ ada $\delta ={\displaystyle \frac{\epsilon }{14}}$ maka $0<|x-1|<\delta$ sedemikian hingga$$\begin{array}{rcl}|\frac{14x^2-20x+6}{x-1}-8|& =& |\frac{(14x-6)(x-1)}{x-1}-8|\\ & =& |(14x-6)-8|\\ & =& |14x-14|\\ & =& |14(x-1)|\\ & =& 14|x-1|\\ & <& 14\delta \\ & =& 14\cdot \frac{\epsilon }{14}=\epsilon \end{array}$$Sehingga terbukti bahwa ${\displaystyle \underset{x\to 1}{lim}\frac{14x^2-20x+6}{x-1}=8}$

2.   Hitunglah limit-limit berikut (jika ada) a. ${\displaystyle \underset{x\to 1}{lim}}{\displaystyle \frac{(3x+4)(2x-2{)}^3}{(x-1{)}^2}}$ b. ${\displaystyle \underset{\theta \to 0}{lim}}{\displaystyle \frac{\tan (5\theta )}{\sin (2\theta )}}$ c. ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}}{\displaystyle \frac{n}{n^2+1}}$ d. ${\displaystyle \underset{x\to 3}{lim}}{\displaystyle \frac{3+x}{3-x}}$

Jawab:

a. ${\displaystyle \underset{x\to 1}{lim}}{\displaystyle \frac{(3x+4)(2x-2{)}^3}{(x-1{)}^2}}$ $$\begin{array}{rcl}{\displaystyle \underset{x\to 1}{lim}}{\displaystyle \frac{(3x+4)(2x-2{)}^3}{(x-1{)}^2}}& =& {\displaystyle \underset{x\to 1}{lim}}{\displaystyle \frac{(3x+4)(2(x-1){)}^3}{(x-1{)}^2}}\\ & =& {\displaystyle \underset{x\to 1}{lim}}{\displaystyle \frac{(3x+4)8(x-1{)}^3}{(x-1{)}^2}}\\ & =& {\displaystyle \underset{x\to 1}{lim}}{\displaystyle 8(3x+4)(x-1)}\\ & =& 0\end{array}$$

b. ${\displaystyle \underset{\theta \to 0}{lim}}{\displaystyle \frac{\tan (5\theta )}{\sin (2\theta )}}$ $$\begin{array}{rcl}{\displaystyle \underset{\theta \to 0}{lim}}{\displaystyle \frac{\tan (5\theta )}{\sin (2\theta )}}& =& {\displaystyle \underset{\theta \to 0}{lim}}{\displaystyle \frac{{\displaystyle \frac{\sin (5\theta )}{\cos (5\theta )}}}{\sin (2\theta )}}\\ & =& {\displaystyle \underset{\theta \to 0}{lim}}{\displaystyle \frac{\sin (5\theta )}{\cos (5\theta )\cdot \sin (2\theta )}}\\ & =& {\displaystyle \underset{\theta \to 0}{lim}}{\displaystyle \frac{1}{\cos (5\theta )}\cdot \frac{\sin (5\theta )}{\sin (2\theta )}}\\ & =& {\displaystyle \underset{\theta \to 0}{lim}}{\displaystyle \frac{1}{\cos (5\theta )}\cdot \sin (5\theta )\cdot \frac{1}{\sin (2\theta )}}\\ & =& {\displaystyle \underset{\theta \to 0}{lim}}{\displaystyle \frac{1}{\cos (5\theta )}\cdot 5\cdot \frac{\sin (5\theta )}{5\theta }\cdot \frac{1}{2}\cdot \frac{2\theta }{\sin (2\theta )}}\\ & =& \frac{5}{2}\underset{\theta \to 0}{lim}\frac{1}{\cos (5\theta )}\cdot \frac{\sin (5\theta )}{5\theta }\cdot \frac{2\theta }{\sin (2\theta )}\\ & =& \frac{5}{2}\cdot 1\cdot 1\cdot 1\\ & =& \frac{5}{2}\end{array}$$

c. ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}}{\displaystyle \frac{n}{n^2+1}}$ $$\begin{array}{rcl}{\displaystyle \underset{n\to \mathrm{\infty }}{lim}}{\displaystyle \frac{n}{n^2+1}}& =& {\displaystyle \underset{n\to \mathrm{\infty }}{lim}}{\displaystyle \frac{{\displaystyle \frac{n}{n^2}}}{{\displaystyle \frac{n^2}{n^2}}+{\displaystyle \frac{1}{n^2}}}}\\ & =& \underset{n\to \mathrm{\infty }}{lim}\frac{{\displaystyle \frac{1}{n}}}{{\displaystyle 1+\frac{1}{n^2}}}\\ & =& \frac{0}{1+0}\\ & =& 0\end{array}$$ d. ${\displaystyle \underset{x\to 3}{lim}}{\displaystyle \frac{3+x}{3-x}}$.Limit tersebut tidak ada. Dengan menggunakan pendekatan limit dibeberapa titik maka kita dapatkan

Terlihat bahwa limit kiri tidak sama dengan limit kanan sehingga $\underset{x\to 3}{lim}f(x)$ tidak ada.

3.  Andaikan $f(x)=\{\begin{array}{ll}-1& \text{jika }x\le 0\\ ax+b& \text{jika }0<x<1\\ 1& \text{jika }x\ge 1\end{array}$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 $$\begin{array}{rcl}a(0)+b& =& -1\\ 0+b& =& -1\\ b& =& -1\end{array}$$Untuk $b=-1$ maka $$\begin{array}{rcl}a(1)+b& =& 1\\ a+(-1)& =& 1\\ a& =& 2\end{array}$$Sehingga kita dapatkan $a=2$ dan $b=-1$.

4.  Dengan menggunakan $f^{\prime }(x)={\displaystyle \underset{h\to 0}{lim}\frac{[f(x+h)-f(x)]}{h}}$ carilah turunan dari $f(x)={\displaystyle \frac{2x-1}{x-4}}$

Jawab:

$$\begin{array}{rcl}f(x)& =& {\displaystyle \frac{2x-1}{x-4}}\\ f^{\prime }(x)& =& \underset{h\to 0}{lim}\frac{[f(x+h)-f(x)]}{h}\\ & =& \underset{h\to 0}{lim}\frac{[{\displaystyle \frac{2(x+h)-1}{(x+h)-4}}-{\displaystyle \frac{2x-1}{x-4}}]}{h}\\ & =& \underset{h\to 0}{lim}\frac{\left[{\displaystyle \frac{(2x+2h-1)(x-4)-(2x-1)(x+h-4)}{(x+h-4)(x-4)}}\right]}{h}\\ & =& \underset{h\to 0}{lim}\left[{\displaystyle \frac{(2x+2h-1)(x-4)-(2x-1)(x+h-4)}{(x+h-4)(x-4)}}\right]\cdot \frac{1}{h}\\ & =& \underset{h\to 0}{lim}\left[{\displaystyle \frac{(2x^2-8x+2xh-8h-x+4)-(2x^2+2xh-8x-x-h+4)}{(x+h-4)(x-4)}}\right]\cdot \frac{1}{h}\\ & =& \underset{h\to 0}{lim}[{\displaystyle \frac{(2x^2-9x+2xh-8h+4)-(2x^2+2xh-9x-h+4)}{(x+h-4)(x-4)}}\cdot \frac{1}{h}]\\ & =& \underset{h\to 0}{lim}[{\displaystyle \frac{(2x^2-9x+2xh-8h+4)-(2x^2+2xh-9x-h+4)}{(x+h-4)(x-4)}}\cdot \frac{1}{h}]\\ & =& \underset{h\to 0}{lim}[\frac{-7h}{(x+h-4)(x-4)}\cdot \frac{1}{h}]\\ & =& \underset{h\to 0}{lim}\left[\frac{-7}{(x+h-4)(x-4)}\right]\\ & =& \left[\frac{-7}{(x-4)(x-4)}\right]\\ & =& -\frac{7}{(x-4{)}^2}\end{array}$$

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

Jawab:

a. $y={\sin }^4(x^2+3x)$ $$\begin{array}{rcl}y& =& {\sin }^4(x^2+3x)\\ y^{\prime }& =& 4{\sin }^3(x^2+3x)\cos (x^2+3x)(2x+3)\\ y^{\prime }& =& (8x+12){\sin }^3(x^2+3x)\cos (x^2+3x)\end{array}$$ b. $x{y}^2+y{x}^2=1$ $$\begin{array}{rcl}x{y}^2+y{x}^2& =& 1\\ y^{2}dx+2xydy+x^{2}dy+2xydx& =& 0\\ dx(y^2+2xy)+dy(2xy+x^2)& =& 0\\ dy(2xy+x^2)& =& -dx(y^2+2xy)\\ \frac{dy}{dx}& =& -\frac{(y^2+2xy)}{(2xy+x^2)}\end{array}$$


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