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

Explanation:

Step1: Rationalize the expression

Multiply and divide by the conjugate (\sqrt{x^{2}+7x + 13}+x) [ \begin{align*} \lim_{x\rightarrow\infty}(\sqrt{x^{2}+7x + 13}-x)&=\lim_{x\rightarrow\infty}\frac{(\sqrt{x^{2}+7x + 13}-x)(\sqrt{x^{2}+7x + 13}+x)}{\sqrt{x^{2}+7x + 13}+x}\ &=\lim_{x\rightarrow\infty}\frac{(x^{2}+7x + 13)-x^{2}}{\sqrt{x^{2}+7x + 13}+x}\ &=\lim_{x\rightarrow\infty}\frac{7x + 13}{\sqrt{x^{2}+7x + 13}+x} \end{align*} ]

Step2: Divide numerator and denominator by (x)

[ \begin{align*} \lim_{x\rightarrow\infty}\frac{7x + 13}{\sqrt{x^{2}+7x + 13}+x}&=\lim_{x\rightarrow\infty}\frac{7+\frac{13}{x}}{\sqrt{1+\frac{7}{x}+\frac{13}{x^{2}}}+1} \end{align*} ]

Step3: Evaluate the limit

As (x\rightarrow\infty), (\frac{13}{x}\rightarrow0), (\frac{7}{x}\rightarrow0) and (\frac{13}{x^{2}}\rightarrow0) [ \begin{align*} \lim_{x\rightarrow\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})