$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+5 x-7}-x\right)=$

$\lim _{x \rightarrow \infty}\left(\sqrt{x^2+5 x-7}-x\right)=$
  1. $\frac{7}{2}$
  2. $5$
  3. $\frac{5}{2}$
  4. $6$

Solution

$\begin{aligned} & \lim _{x \rightarrow \infty} \sqrt{x^2+5-7}-x \\ & =\lim _{x \rightarrow \infty} \frac{\left(\sqrt{x^2+5 x-7}\right)\left(\sqrt{x^2+5 x-7+x}\right)}{\left(\sqrt{x^2+5 x-7+x}\right)} \\ & =\lim _{x \rightarrow \infty} \frac{x^2+5 x-7-x^2}{\left(\sqrt{x^2+5 x-7+x}\right)} \end{aligned}$ Dividing numerator and denominator by $\mathrm{x}$, we get $=\lim _{x \rightarrow \infty} \frac{5-\frac{7}{x}}{\left(\sqrt{1+\frac{5}{x}-\frac{7}{x^2}+1}\right)}=\frac{5}{\sqrt{1}+1}=\frac{5}{2}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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