Question

1. 0 / 3 points evaluate the limit. $lim_{x

ightarrowinfty}(sqrt{x^{2}+9x}-x)$

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:

$\frac{9}{2}$