$$ lim _ { x
ightarrow infty } sqrt { x ^ { 2 } + 7 x + 13 } - x $$

Question

Accepted Answer

# Explanation:
## Step1: Rationalize the expression
Multiply and divide by the conjugate $\sqrt{x^{2}+7x + 13}+x$
$$
\begin{align*}
\lim_{x
ightarrow\infty}(\sqrt{x^{2}+7x + 13}-x)&=\lim_{x
ightarrow\infty}\frac{(\sqrt{x^{2}+7x + 13}-x)(\sqrt{x^{2}+7x + 13}+x)}{\sqrt{x^{2}+7x + 13}+x}\
&=\lim_{x
ightarrow\infty}\frac{(x^{2}+7x + 13)-x^{2}}{\sqrt{x^{2}+7x + 13}+x}\
&=\lim_{x
ightarrow\infty}\frac{7x + 13}{\sqrt{x^{2}+7x + 13}+x}
\end{align*}
$$
## Step2: Divide numerator and denominator by $x$
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{7x + 13}{\sqrt{x^{2}+7x + 13}+x}&=\lim_{x
ightarrow\infty}\frac{7+\frac{13}{x}}{\sqrt{1+\frac{7}{x}+\frac{13}{x^{2}}}+1}
\end{align*}
$$
## Step3: Evaluate the limit
As $x
ightarrow\infty$, $\frac{13}{x}
ightarrow0$, $\frac{7}{x}
ightarrow0$ and $\frac{13}{x^{2}}
ightarrow0$
$$
\begin{align*}
\lim_{x
ightarrow\infty}\frac{7+\frac{13}{x}}{\sqrt{1+\frac{7}{x}+\frac{13}{x^{2}}}+1}&=\frac{7 + 0}{\sqrt{1+0+0}+1}\
&=\frac{7}{2}
\end{align*}
$$

# Answer:
$\frac{7}{2}$

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QUESTION IMAGE

Question

Explanation:

Step1: Rationalize the expression

Step2: Divide numerator and denominator by \(x\)

Step3: Evaluate the limit

Answer: