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