5. 0 / 3 points evaluate the limit. $lim_{x
ightarrowinfty}(sqrt{x^{2}+9x}-x)$

Question

Accepted Answer

# Explanation:
## Step1: Rationalize the expression
Multiply and divide by $\sqrt{x^{2}+9x}+x$.
$$
\begin{align*}
&\lim_{x
ightarrow\infty}(\sqrt{x^{2}+9x}-x)\times\frac{\sqrt{x^{2}+9x}+x}{\sqrt{x^{2}+9x}+x}\
=&\lim_{x
ightarrow\infty}\frac{(\sqrt{x^{2}+9x}-x)(\sqrt{x^{2}+9x}+x)}{\sqrt{x^{2}+9x}+x}\
=&\lim_{x
ightarrow\infty}\frac{(x^{2}+9x)-x^{2}}{\sqrt{x^{2}+9x}+x}\
=&\lim_{x
ightarrow\infty}\frac{9x}{\sqrt{x^{2}+9x}+x}
\end{align*}
$$
## Step2: Divide numerator and denominator by $x$
Since $x
ightarrow\infty$, we have:
$$
\begin{align*}
&\lim_{x
ightarrow\infty}\frac{9x/x}{(\sqrt{x^{2}+9x}+x)/x}\
=&\lim_{x
ightarrow\infty}\frac{9}{\sqrt{1 + \frac{9}{x}}+1}
\end{align*}
$$
## Step3: Evaluate the limit
As $x
ightarrow\infty$, $\frac{9}{x}
ightarrow0$.
$$
\begin{align*}
&\lim_{x
ightarrow\infty}\frac{9}{\sqrt{1 + \frac{9}{x}}+1}=\frac{9}{\sqrt{1+0}+1}\
=&\frac{9}{2}
\end{align*}
$$
# Answer:
$\frac{9}{2}$

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QUESTION IMAGE

Question

Explanation:

Step1: Rationalize the expression

Step2: Divide numerator and denominator by $x$

Step3: Evaluate the limit

Answer: