$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=$
$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}=$
- $\frac{1}{5}$
- $\frac{1}{10}$
- $\frac{-1}{10}$
- $\frac{-1}{5}$
Solution
$\begin{aligned}
& \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} \\
& \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(x-1)(2 x+3)}=\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)(2 x+3)} \\
& \lim _{x \rightarrow 1} \frac{(2 x-3)}{(\sqrt{x}+1)(2 x+3)}=\frac{-1}{2(5)}=\frac{-1}{10}
\end{aligned}$
Asked in: MHT CET 2021 (21 Sep Shift 1)
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