Question

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

• A. $\frac{1}{10}$
• B. $-\frac{1}{10}$
• C. $1$
• D. none of these

Answer 1

Answer: (B)

$-\frac{1}{10}$

${lim}_{x\to 1}\frac{(\sqrt{x}-1)(2x-3)}{2x^2+x-3}={lim}_{x\to 1}\frac{(\sqrt{x}-1)(2x-3)}{(x-1)(2x+3)}\frac{(\sqrt{x}+1)}{(\sqrt{x}+1)}={lim}_{x\to 1}\frac{(x-1)(2x-3)}{(x-1)(2x+3)(\sqrt{x}+1)}={lim}_{x\to 1}\frac{(2x-3)}{(2x+3)(\sqrt{x}+1)}=-\frac{1}{10}$
Hence, option 'B' is correct.