Question

1. show algebraically that $lim_{x

ightarrowinfty}sqrt{x^{2}-2}-x = 0$.

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.