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Q. $\displaystyle \lim_{x \to 1}$ $\frac{\left(\sqrt{x}-1\right)\left(2x-3\right)}{2x^{2}+x-3}$ is

Limits and Derivatives

Solution:

We have, $\displaystyle \lim_{x \to 1}$ $\frac{\left(\sqrt{x}-1\right)\left(2x-3\right)}{2x^{2}+x-3}$
$=\displaystyle \lim_{x \to 1}$ $\frac{\left(\sqrt{x}-1\right)\left(2x-3\right)}{\left(x-1\right)\left(2x+3\right)}$
$=\displaystyle \lim_{x \to 1}$ $\frac{\left(2x-3\right)}{\left(2x+3\right)} \times \displaystyle \lim_{x \to 1}$ $\left(\frac{x^{\frac{1}{2}}-1^{\frac{1}{2}}}{x-1}\right)$
$=\frac{-1}{5} \times\frac{1}{2}\left(1\right)^{-\frac{1}{2}}$
$=\frac{-1}{5}\times\frac{1}{2}=\frac{-1}{10}$