Question

$\underset{x\to 1}{lim}\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{2x^2+x-3}=$

• A. $\frac{1}{10}$
• B. $\frac{-1}{10}$
• C. $\frac{2}{5}$
• D. $-\frac{2}{5}$

Answer 1

Answer: (B)

$\frac{-1}{10}$

Now,

$\underset{x\to 1}{lim}\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{2x^2+x-3}$

$=\underset{x\to 1}{lim}\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{(2x^2+3x-2x-3)}$

$=\underset{x\to 1}{lim}\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{(x-1)(2x+3)}$

$=\underset{x\to 1}{lim}\frac{\left(2x-3\right)\left(\sqrt{x}-1\right)}{(\sqrt{x}-\sqrt{1})(\sqrt{x}+\sqrt{1})(2x+3)}$

$=\underset{x\to 1}{lim}\frac{\left(2x-3\right)}{(\sqrt{x}+\sqrt{1})(2x+3)}$

$=\frac{-1}{2\times 5}$

$=-\frac{1}{10}$.