[tex]\lim _ { x \rightarrow \infty } x ( \sqrt { x ^ { 2 } + 1 } - x )[/tex]
beserta caranya yaa, terima kasih!
Jawab:
[tex]\large\text{$\begin{aligned}&{\lim_{x\to\infty}}\:x\!\left(\sqrt{x^2+1}-x\right)\ =\ \bf\frac{1}{2}\end{aligned}$}[/tex]
Penjelasan dengan langkah-langkah:
Limit
[tex]\large\text{$\begin{aligned}&{\lim_{x\to\infty}}\:x\!\left(\sqrt{x^2+1}-x\right)\\\\&{=\ }{\lim_{x\to\infty}}\:\frac{x\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)}{\qquad\qquad\qquad\quad\,\left(\sqrt{x^2+1}+x\right)}\\\\&{=\ }{\lim_{x\to\infty}}\:\frac{x(x^2+1-x^2)}{\sqrt{x^2+1}+x}\\\\&{=\ }{\lim_{x\to\infty}}\:\frac{x(1)}{\sqrt{x^2+1}+x}\ =\ {\lim_{x\to\infty}}\:\frac{x}{\sqrt{x^2+1}+x}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&{=\ }{\lim_{x\to\infty}}\:\frac{\qquad\ \ x\quad\qquad\,\left(\frac{1}{x}\right)}{\left(\sqrt{x^2+1}+x\right)\left(\frac{1}{x}\right)}\\\\&{=\ }{\lim_{x\to\infty}}\:\frac{1}{\frac{\sqrt{x^2+1}}{x}+1}\ =\ {\lim_{x\to\infty}}\:\frac{1}{\sqrt{\frac{x^2+1}{x^2}}+1}\\\\&{=\ }{\lim_{x\to\infty}}\:\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&{\quad}\left[\ \normalsize\text{$\begin{aligned}&\textsf{Dengan pengecualian pada bentuk tak tentu,}\\&\textsf{berlaku $\lim _{x\to a}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x\to a}f(x)}{\lim\limits_{x\to a}g(x)},\quad\lim_{x\to a}g(x)\ne0$}\end{aligned}$}\right.\\\\&{=\ }\frac{{\lim\limits_{x\to\infty}}\:1}{\lim\limits_{x\to\infty}\:\left(\sqrt{1+\frac{1}{x^2}}+1\right)}\\\\&{=\ }\frac{1}{\lim\limits_{x\to\infty}\:\left(\sqrt{1+\frac{1}{x^2}}+1 \right)}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&{\quad}\left[\ \normalsize\text{$\begin{aligned}&\textsf{Dengan pengecualian pada bentuk tak tentu,}\\&\textsf{berlaku $\lim_{x\to a}\left[f(x)\pm g(x)\right]=\lim\limits_{x\to a}f(x)\pm\lim\limits_{x\to a}g(x)$}\end{aligned}$}\right.\\\\&{=\ }\frac{1}{\lim\limits_{x\to\infty}\:\sqrt{1+\frac{1}{x^2}}+\lim\limits_{x\to\infty}\:1}\\\\&{=\ }\frac{1}{\left(\lim\limits_{x\to\infty}\:\sqrt{1+\frac{1}{x^2}}\right)+1}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&{\quad}\left[\ \normalsize\text{$\begin{aligned}&\textsf{Dengan pengecualian pada bentuk tak tentu,}\\&\textsf{berlaku $\lim_{x\to a}\left[f(x)\right]^b=\left[\lim_{x\to a}f(x)\right]^b$}\end{aligned}$}\right.\\\\&{=\ }\frac{1}{\sqrt{\lim\limits_{x\to\infty}\:\left(1+\frac{1}{x^2}\right)}+1}\\\\&{=\ }\frac{1}{\sqrt{\lim\limits_{x\to\infty}\:1+\lim\limits_{x\to\infty}\:\frac{1}{x^2}}+1}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}&{\quad}\left[\ \normalsize\text{$\begin{aligned}&\lim_{x\to\infty}\frac{c}{x^a}=0\end{aligned}$}\right.\\\\&{=\ }\frac{1}{\sqrt{1+0}+1}\ =\ \frac{1}{\sqrt{1}+1}\ =\ \frac{1}{1+1}\\\\&{=\ }\bf\frac{1}{2}\end{aligned}$}[/tex]
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