MHT CET · Maths · Limits
\(
\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=
\)
- A \(\frac{1}{5}\)
- B \(\frac{1}{10}\)
- C \(\frac{-1}{10}\)
- D \(\frac{-1}{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{-1}{10}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} \\
& \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(x-1)(2 x+3)}=\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)(2 x+3)} \\
& \lim _{x \rightarrow 1} \frac{(2 x-3)}{(\sqrt{x}+1)(2 x+3)}=\frac{-1}{2(5)}=\frac{-1}{10}
\end{aligned}
\)
\begin{aligned}
& \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} \\
& \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(x-1)(2 x+3)}=\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)(2 x+3)} \\
& \lim _{x \rightarrow 1} \frac{(2 x-3)}{(\sqrt{x}+1)(2 x+3)}=\frac{-1}{2(5)}=\frac{-1}{10}
\end{aligned}
\)
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