Question

evaluate the following limits. if needed, enter inf for ∞ and minf for -∞.
(a)
\\(\lim_{x\to\infty}(\sqrt{x^{2}+5x + 1}-x)=\\)
(b)
\\(\lim_{x\to -\infty}(\sqrt{x^{2}+5x + 1}-x)=\\)

Explanation:

Step1: Rationalize the expression

Multiply and divide by $\sqrt{x^{2}+5x + 1}+x$.
\[

$$\begin{align*} \lim_{x ightarrow\infty}(\sqrt{x^{2}+5x + 1}-x)&=\lim_{x ightarrow\infty}\frac{(\sqrt{x^{2}+5x + 1}-x)(\sqrt{x^{2}+5x + 1}+x)}{\sqrt{x^{2}+5x + 1}+x}\\ &=\lim_{x ightarrow\infty}\frac{(x^{2}+5x + 1)-x^{2}}{\sqrt{x^{2}+5x + 1}+x}\\ &=\lim_{x ightarrow\infty}\frac{5x + 1}{\sqrt{x^{2}+5x + 1}+x} \end{align*}$$

\]

Step2: Divide numerator and denominator by $x$

As $x
ightarrow\infty$, $x>0$, so $\sqrt{x^{2}}=x$.
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{5x + 1}{\sqrt{x^{2}+5x + 1}+x}&=\lim_{x ightarrow\infty}\frac{5+\frac{1}{x}}{\sqrt{1+\frac{5}{x}+\frac{1}{x^{2}}}+1}\\ \end{align*}$$

\]

Step3: Evaluate the limit

Using the limit rules $\lim_{x
ightarrow\infty}\frac{1}{x}=0$ and $\lim_{x
ightarrow\infty}\frac{1}{x^{2}} = 0$.
\[

$$\begin{align*} \lim_{x ightarrow\infty}\frac{5+\frac{1}{x}}{\sqrt{1+\frac{5}{x}+\frac{1}{x^{2}}}+1}&=\frac{5 + 0}{\sqrt{1+0 + 0}+1}\\ &=\frac{5}{2} \end{align*}$$

\]

Step4: For $x

ightarrow-\infty$
Multiply and divide by $\sqrt{x^{2}+5x + 1}+x$.
\[

$$\begin{align*} \lim_{x ightarrow-\infty}(\sqrt{x^{2}+5x + 1}-x)&=\lim_{x ightarrow-\infty}\frac{(x^{2}+5x + 1)-x^{2}}{\sqrt{x^{2}+5x + 1}+x}\\ &=\lim_{x ightarrow-\infty}\frac{5x + 1}{\sqrt{x^{2}+5x + 1}+x} \end{align*}$$

\]
Since $x
ightarrow-\infty$, $\sqrt{x^{2}}=-x$. Divide numerator and denominator by $-x$.
\[

$$\begin{align*} \lim_{x ightarrow-\infty}\frac{5x + 1}{\sqrt{x^{2}+5x + 1}+x}&=\lim_{x ightarrow-\infty}\frac{-5-\frac{1}{x}}{\sqrt{1+\frac{5}{x}+\frac{1}{x^{2}}}-1}\\ &=\frac{- 5-0}{\sqrt{1+0 + 0}-1}=\text{INF} \end{align*}$$

\]

Answer:

(a) $\frac{5}{2}$
(b) INF