evaluate the following limits. if needed, enter inf for ∞ and minf for -∞.\n(a)\n\\(\\lim_{x\\to\\infty}(\\sq…

Explanation:

Step1: Rationalize the expression

Multiply and divide by $\sqrt{x^{2}+5x + 1}+x$. [ \begin{align*} \lim_{x\rightarrow\infty}(\sqrt{x^{2}+5x + 1}-x)&=\lim_{x\rightarrow\infty}\frac{(\sqrt{x^{2}+5x + 1}-x)(\sqrt{x^{2}+5x + 1}+x)}{\sqrt{x^{2}+5x + 1}+x}\ &=\lim_{x\rightarrow\infty}\frac{(x^{2}+5x + 1)-x^{2}}{\sqrt{x^{2}+5x + 1}+x}\ &=\lim_{x\rightarrow\infty}\frac{5x + 1}{\sqrt{x^{2}+5x + 1}+x} \end{align*} ]

Step2: Divide numerator and denominator by $x$

As $x\rightarrow\infty$, $x>0$, so $\sqrt{x^{2}}=x$. [ \begin{align*} \lim_{x\rightarrow\infty}\frac{5x + 1}{\sqrt{x^{2}+5x + 1}+x}&=\lim_{x\rightarrow\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 $\lim_{x\rightarrow\infty}\frac{1}{x}=0$ and $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$. [ \begin{align*} \lim_{x\rightarrow\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 $x\rightarrow-\infty$

Multiply and divide by $\sqrt{x^{2}+5x + 1}+x$. [ \begin{align*} \lim_{x\rightarrow-\infty}(\sqrt{x^{2}+5x + 1}-x)&=\lim_{x\rightarrow-\infty}\frac{(x^{2}+5x + 1)-x^{2}}{\sqrt{x^{2}+5x + 1}+x}\ &=\lim_{x\rightarrow-\infty}\frac{5x + 1}{\sqrt{x^{2}+5x + 1}+x} \end{align*} ] Since $x\rightarrow-\infty$, $\sqrt{x^{2}}=-x$. Divide numerator and denominator by $-x$. [ \begin{align*} \lim_{x\rightarrow-\infty}\frac{5x + 1}{\sqrt{x^{2}+5x + 1}+x}&=\lim_{x\rightarrow-\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