Berikut ini adalah kumpulan soal mengenai
. Jika ingin bertanya soal, silahkan gabung ke grup
Telegram,
Signal,
Discord, atau
WhatsApp.
No.
Buktikan limit fungsi berikut secara definisi
\displaystyle\lim_{(x,y)\to(3,2)}(3x-4y)=1
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}
No.
Diketahui
{f(x)=x^2-1}. Tentukan nilai dari
\displaystyle\lim_{p\to0}\dfrac{f(2+p)-f(2)}p
\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\\
&=\displaystyle\lim_{p\to0}\dfrac{4+4p+p^2-1-3}p\\
&=\displaystyle\lim_{p\to0}\dfrac{4p+p^2}p\\
&=\displaystyle\lim_{p\to0}(4+p)\\
&=4+0\\
&=\boxed{\boxed{4}}
\end{aligned}
No.
\displaystyle\lim_{x\to0}\dfrac{x}{\sqrt{5-x}-\sqrt5}= ....
\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}
No.
\displaystyle\lim_{x\to0}\dfrac{x^3+2x^2}{x^4-x^3+5x^2}= ....
\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}
No.
\displaystyle\lim_{x\to-3}\dfrac{2x+1}{x^2-3x+4}= ....
\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}
No.
\displaystyle\lim_{x\to0}\dfrac{x^2+3x-1}{x^3+5x}= ....
\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}
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?
| 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 |
No.
\displaystyle\lim_{x\to3}x^2+2x+10= ....
\begin{aligned}
\displaystyle\lim_{x\to3}x^2+2x+10&=3^2+2(3)+10\\
&=9+6+10\\
&=\boxed{\boxed{25}}
\end{aligned}
0 Komentar
Silahkan berkomentar dengan santun di sini. Anda juga boleh bertanya soal matematika atau mengoreksi jawaban di atas