Exercise Zone : Limit [3]

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Tipe:

Standar SBMPTN HOTS

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No.

Buktikan limit fungsi berikut secara definisi
\displaystyle\lim_{(x,y)\to(3,2)}(3x-4y)=1

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Cari \delta\lt0 sedemikian rupa sehingga
{0\lt\sqrt{(x-3)^2+(y-2)^2}\lt\delta\Rightarrow|(3x-4y)-1|\lt\epsilon}

\sqrt{(x-3)^2+(y-2)^2}\lt\delta didapat |x-3|\lt\delta dan |y-2|\lt\delta. Ambil \delta=\dfrac{\epsilon}7

{|3x-4y-1|\lt|3x-9-4y+8|\lt3|x-3|+4|y-2|\lt3\delta+4\delta=7\delta=7\left(\dfrac{\epsilon}7\right)=\epsilon}

Jadi, terbukti bahwa \displaystyle\lim_{(x,y)\to(3,2)}(3x-4y)=1.

No.

Diketahui {f(x)=x^2-1}. Tentukan nilai dari \displaystyle\lim_{p\to0}\dfrac{f(2+p)-f(2)}p

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$$\begin{array}{rl}{\displaystyle \underset{p\to 0}{lim}{\displaystyle \frac{f(2+p)-f(2)}{p}}}& ={\displaystyle \underset{p\to 0}{lim}{\displaystyle \frac{(2+p{)}^2-1-(2^2-1)}{p}}}\\ & ={\displaystyle \underset{p\to 0}{lim}{\displaystyle \frac{4+4p+p^2-1-3}{p}}}\\ & ={\displaystyle \underset{p\to 0}{lim}{\displaystyle \frac{4p+p^2}{p}}}\\ & ={\displaystyle \underset{p\to 0}{lim}(4+p)}\\ & =4+0\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$$

Jadi, \displaystyle\lim_{p\to0}\dfrac{f(2+p)-f(2)}p=4.

No.

\displaystyle\lim_{x\to0}\dfrac{x}{\sqrt{5-x}-\sqrt5}= ....

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$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x}{\sqrt{5-x}-\sqrt{5}}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x}{\sqrt{5-x}-\sqrt{5}}}\cdot {\displaystyle \frac{\sqrt{5-x}+\sqrt{5}}{\sqrt{5-x}+\sqrt{5}}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x(\sqrt{5-x}+\sqrt{5})}{5-x-5}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x(\sqrt{5-x}+\sqrt{5})}{-x}}}\\ & ={\displaystyle \underset{x\to 0}{lim}(-\sqrt{5-x}-\sqrt{5})}\\ & =-\sqrt{5-0}-\sqrt{5}\\ & =-\sqrt{5}-\sqrt{5}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -2\sqrt{5}}}}}\end{array}$$

Jadi, \displaystyle\lim_{x\to0}\dfrac{x}{\sqrt{5-x}-\sqrt5}=-2\sqrt5.

No.

\displaystyle\lim_{x\to0}\dfrac{x^3+2x^2}{x^4-x^3+5x^2}= ....

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$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x^3+2x^2}{x^4-x^3+5x^2}}}& ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x^2(x+2)}{x^2(x^2-x+5)}}}\\ & ={\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x+2}{x^2-x+5}}}\\ & ={\displaystyle \frac{0+2}{0^2-0+5}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2}{5}}}}}}\end{array}$$

Jadi, \displaystyle\lim_{x\to0}\dfrac{x^3+2x^2}{x^4-x^3+5x^2}=\dfrac25.

No.

\displaystyle\lim_{x\to-3}\dfrac{2x+1}{x^2-3x+4}= ....

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$$\begin{array}{rl}{\displaystyle \underset{x\to -3}{lim}{\displaystyle \frac{2x+1}{x^2-3x+4}}}& ={\displaystyle \frac{2(-3)+1}{(-3{)}^2-3(-3)+4}}\\ & ={\displaystyle \frac{-6+1}{9+9+4}}\\ & ={\displaystyle \frac{-5}{22}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{5}{22}}}}}}\end{array}$$

Jadi, \displaystyle\lim_{x\to-3}\dfrac{2x+1}{x^2-3x+4}=-\dfrac5{22}.

No.

\displaystyle\lim_{x\to0}\dfrac{x^2+3x-1}{x^3+5x}= ....

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$$\begin{array}{rl}{\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x^2+3x-1}{x^3+5x}}}& ={\displaystyle \frac{0^2+3(0)-1}{0^3+5(0)}}\\ & ={\displaystyle \frac{-1}{0}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \mathrm{\infty }}}}}\end{array}$$

Jadi, \displaystyle\lim_{x\to0}\dfrac{x^2+3x-1}{x^3+5x}=\infty.

No.

\displaystyle\lim_{x\to2}\dfrac{x^2-x-2}{x-2}
Dengan menggunakan tabel berikut

Nilai x	Nilai y
x=1{,}95
x=1{,}98
x=1{,}99
x=2{,}01
x=2{,}05
x=2{,}1

Berapakah hasil yang kamu peroleh untuk limit fungsi tersebut?

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Nilai x	Nilai y
x=1{,}95	2{,}95
x=1{,}98	2{,}98
x=1{,}99	2{,}99
x=2{,}01	3{,}01
x=2{,}05	3{,}05
x=2{,}1	3{,}1

Jadi, \displaystyle\lim_{x\to2}\dfrac{x^2-x-2}{x-2}=3.

No.

\displaystyle\lim_{x\to3}x^2+2x+10= ....

$$\begin{array}{rl}{\displaystyle \underset{x\to 3}{lim}{x}^2+2x+10}& =3^2+2(3)+10\\ & =9+6+10\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 25}}}}\end{array}$$

Jadi, \displaystyle\lim_{x\to3}x^2+2x+10=25.