Soal-Soal Tentang Limit Fungsi

Berikut ini saya mencoba memposting soal-soal tentang Limit fungsi. Sederhana saja kok soalnya. Mudah-mudahan dapat digunakan sebagai bahan latihan

Nilai dari ${\displaystyle \underset{x\to 2}{lim}\frac{x^2-4}{x^2-2x}=\cdots }$

Pembahasan :
$$\begin{array}{rcl}\underset{x\to 2}{lim}\frac{x^2-4}{x^2-2x}& =& \underset{x\to 2}{lim}\frac{(x-2)(x+2)}{x(x-2)}\\ & =& \underset{x\to 2}{lim}\frac{x+2}{x}\\ & =& \frac{4}{2}\\ & =& 2\end{array}$$

${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\frac{3x^2-2x+1}{{(2x-3)}^2}=\cdots }$

Pembahasan :

\begin{eqnarray*}\lim_{x\to\infty}\frac{3x^{2}-2x+1}{\left(2x-3\right)^{2}} & = & \lim_{x\to\infty}\frac{3x^{2}-2x+1}{4x^{2}-12x+9}\
& = & \lim_{x\to\infty}\frac{3-\frac{2}{x}+\frac{1}{x^{2}}}{4-\frac{12}{x}+\frac{9}{x^{2}}}\
& = & \frac{3-0+0}{4-0+0}\
& = & \frac{3}{4}
\end{eqnarray*}

${\displaystyle \underset{x\to \mathrm{\infty }}{lim}(\sqrt{x^2+3x-1}-\sqrt{x^2-5x+2})=\cdots }$

Pembahasan :

\begin{eqnarray*}
& = & \lim_{x\to\infty}\left(\sqrt{x^{2}+3x-1}-\sqrt{x^{2}-5x+2}\right)\times\frac{\left(\sqrt{x^{2}+3x-1}+\sqrt{x^{2}-5x+2}\right)}{\left(\sqrt{x^{2}+3x-1}+\sqrt{x^{2}-5x+2}\right)}\

& = & \lim_{x\to\infty}\frac{\left(x^{2}+3x-1-\left(x^{2}-5x+2\right)\right)}{\sqrt{x^{2}+3x-1}+\sqrt{x^{2}-5x+2}}\
& = & \lim_{x\to\infty}\frac{\left(x^{2}+3x-1-x^{2}+5x-2\right)}{\sqrt{x^{2}+3x-1}+\sqrt{x^{2}-5x+2}}\
& = & \lim_{x\to\infty}\frac{\left(8x-3\right)}{\sqrt{x^{2}+3x-1}+\sqrt{x^{2}-5x+2}}\
& = & \lim_{x\to\infty}\frac{\left(8-\frac{3}{x}\right)}{\sqrt{1+\frac{3}{x}-\frac{1}{x^{2}}}+\sqrt{1-\frac{5}{x}+\frac{2}{x^{2}}}}\
& = & \frac{8-0}{\sqrt{1+0-0}+\sqrt{1-0+0}}\
& = & \frac{8}{2}\
& = & 4
\end{eqnarray*}