Exercise Zone : Limit [3]
Table of Contents
Berikut ini adalah kumpulan soal mengenai Limit. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.
Tipe:
No.
Buktikan limit fungsi berikut secara definisiALTERNATIF PENYELESAIAN
Cari δ < 0 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| < δ dan |y − 2| < δ . 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.
DiketahuiALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{p\to0}\dfrac{f(2+p)-f(2)}p&=\displaystyle\lim_{p\to0}\dfrac{(2+p)^2-1-\left(2^2-1\right)}p\\[3.8pt]
&=\displaystyle\lim_{p\to0}\dfrac{4+4p+p^2-1-3}p\\[3.8pt]
&=\displaystyle\lim_{p\to0}\dfrac{4p+p^2}p\\[3.8pt]
&=\displaystyle\lim_{p\to0}(4+p)\\
&=4+0\\
&=\boxed{\boxed{4}}
\end{aligned}
Jadi, \displaystyle\lim_{p\to0}\dfrac{f(2+p)-f(2)}p=4 .
No.
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to0}\dfrac{x}{\sqrt{5-x}-\sqrt5}&=\displaystyle\lim_{x\to0}\dfrac{x}{\sqrt{5-x}-\sqrt5}{\color{red}\cdot\dfrac{\sqrt{5-x}+\sqrt5}{\sqrt{5-x}+\sqrt5}}\\
&=\displaystyle\lim_{x\to0}\dfrac{x\left(\sqrt{5-x}+\sqrt5\right)}{5-x-5}\\
&=\displaystyle\lim_{x\to0}\dfrac{x\left(\sqrt{5-x}+\sqrt5\right)}{-x}\\
&=\displaystyle\lim_{x\to0}\left(-\sqrt{5-x}-\sqrt5\right)\\
&=-\sqrt{5-0}-\sqrt5\\
&=-\sqrt5-\sqrt5\\
&=\boxed{\boxed{-2\sqrt5}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to0}\dfrac{x}{\sqrt{5-x}-\sqrt5}=-2\sqrt5 .
No.
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to0}\dfrac{x^3+2x^2}{x^4-x^3+5x^2}&=\displaystyle\lim_{x\to0}\dfrac{x^2(x+2)}{x^2\left(x^2-x+5\right)}\\
&=\displaystyle\lim_{x\to0}\dfrac{x+2}{x^2-x+5}\\
&=\dfrac{0+2}{0^2-0+5}\\
&=\boxed{\boxed{\dfrac25}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to0}\dfrac{x^3+2x^2}{x^4-x^3+5x^2}=\dfrac25 .
No.
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to-3}\dfrac{2x+1}{x^2-3x+4}&=\dfrac{2(-3)+1}{(-3)^2-3(-3)+4}\\
&=\dfrac{-6+1}{9+9+4}\\
&=\dfrac{-5}{22}\\
&=\boxed{\boxed{-\dfrac5{22}}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to-3}\dfrac{2x+1}{x^2-3x+4}=-\dfrac5{22} .
No.
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to0}\dfrac{x^2+3x-1}{x^3+5x}&=\dfrac{0^2+3(0)-1}{0^3+5(0)}\\
&=\dfrac{-1}0\\
&=\boxed{\boxed{\infty}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to0}\dfrac{x^2+3x-1}{x^3+5x}=\infty .
No.
Dengan menggunakan tabel berikut
| Nilai x | Nilai y |
ALTERNATIF PENYELESAIAN
| Nilai x | Nilai y |
| 2,95 | |
| 2,98 | |
| 2,99 | |
| 3,01 | |
| 3,05 | |
| 3,1 |
Jadi, \displaystyle\lim_{x\to2}\dfrac{x^2-x-2}{x-2}=3 .
No.
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to3}x^2+2x+10&=3^2+2(3)+10\\
&=9+6+10\\
&=\boxed{\boxed{25}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to3}x^2+2x+10=25 .
No.
- 4
- 5
- 6
- 7
- 8
ALTERNATIF PENYELESAIAN
\begin{aligned}
\displaystyle\lim_{x\to2}(3x+2)&=3(2)+2\\
&=6+2\\
&=\boxed{\boxed{8}}
\end{aligned}
Jadi, \displaystyle\lim_{x\to2}(3x+2)=8 .
JAWAB: E
JAWAB: E
No.
Tentukan definisi limit berikut apakah limitnya "terdefinisi, tidak terdefinisi, nilai limit\displaystyle\lim_{x\to3}\dfrac{x^2-1}{x+1} = ....\displaystyle\lim_{x\to2}\dfrac{2x^2+1}{2x-4} = ....\displaystyle\lim_{x\to4}\dfrac{x^2-16}{x-4} = ....
ALTERNATIF PENYELESAIAN
\begin{aligned} \lim_{x\to3}\frac{x^2-1}{x+1}&=\frac{3^2-1}{3+1}\\[3.7pt] &=\frac{9-1}4\\[3.7pt] &=\frac84\\[3.7pt] &=2 \end{aligned} terdefinisi
\begin{aligned} \lim_{x\to2}\frac{2x^2+1}{2x-4}&=\frac{2(2)^2+1}{2(2)-4}\\[3.7pt] &=\frac{2(4)+1}{4-4}\\[3.7pt] &=\frac90 \end{aligned} tidak terdefinisi
\begin{aligned} \lim_{x\to4}\frac{x^2-16}{x-4}&=\frac{4^2-16}{4-4}\\[3.7pt] &=\frac{16-16}0\\[3.7pt] &=\frac00 \end{aligned} nilai limit\dfrac00
Jadi,
- terdefinisi
- tidak terdefinisi
- nilai limit
\dfrac00
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