MHT CET · Maths · Limits
\(\lim _{x \rightarrow \infty}\left(\sqrt{x^2+5 x-7}-x\right)=\)
- A \(\frac{7}{2}\)
- B \(5\)
- C \(\frac{5}{2}\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \lim _{x \rightarrow \infty} \sqrt{x^2+5-7}-x \\
& =\lim _{x \rightarrow \infty} \frac{\left(\sqrt{x^2+5 x-7}\right)\left(\sqrt{x^2+5 x-7+x}\right)}{\left(\sqrt{x^2+5 x-7+x}\right)} \\
& =\lim _{x \rightarrow \infty} \frac{x^2+5 x-7-x^2}{\left(\sqrt{x^2+5 x-7+x}\right)}
\end{aligned}
\)
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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