3. show algebraically that $lim_{x
ightarrowinfty}sqrt{x^{2}-2}-x = 0$.

Question

Accepted Answer

# Explanation:
## Step1: Rationalize the expression
Multiply and divide by the conjugate $\sqrt{x^{2}-2}+x$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}(\sqrt{x^{2}-2}-x)&=\lim_{x
ightarrow\infty}\frac{(\sqrt{x^{2}-2}-x)(\sqrt{x^{2}-2}+x)}{\sqrt{x^{2}-2}+x}\
&=\lim_{x
ightarrow\infty}\frac{(x^{2}-2)-x^{2}}{\sqrt{x^{2}-2}+x}
\end{align*}
$$
## Step2: Simplify the numerator
Simplify the numerator $(x^{2}-2)-x^{2}$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{(x^{2}-2)-x^{2}}{\sqrt{x^{2}-2}+x}&=\lim_{x
ightarrow\infty}\frac{- 2}{\sqrt{x^{2}-2}+x}
\end{align*}
$$
## Step3: Analyze the limit as $x
ightarrow\infty$
As $x
ightarrow\infty$, $\sqrt{x^{2}-2}\sim x$.
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{-2}{\sqrt{x^{2}-2}+x}&=\lim_{x
ightarrow\infty}\frac{-2}{x + x}\
&=\lim_{x
ightarrow\infty}\frac{-2}{2x}\
&=\lim_{x
ightarrow\infty}\frac{-1}{x}=0
\end{align*}
$$
# Answer:
The limit $\lim_{x
ightarrow\infty}\sqrt{x^{2}-2}-x = 0$ is shown.

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QUESTION IMAGE

Question

Explanation:

Step1: Rationalize the expression

Step2: Simplify the numerator

Step3: Analyze the limit as $x

Answer: