Question

Evaluate $\underset{x\to 2}{lim}\frac{x^{\sqrt{2}}-2^{\sqrt{2}}}{x-2}$

• A. $\sqrt{2}.2^{\sqrt{2}}$
• B. $2^{\sqrt{2-1}}$
• C. $2^{\sqrt{2}-\frac{1}{2}}$
• D. $2^{\sqrt{2}}$

Answer 1

The correct option is C $2^{\sqrt{2}-\frac{1}{2}}$

$\underset{x\to 2}{lim}\frac{x^{\sqrt{2}}-2^{\sqrt{2}}}{x-2}$ $(÷form)$

using L-hospital rule

$=\underset{x\to 2}{lim}\sqrt{2}{x}^{\sqrt{2}-1}$

$=\sqrt{2}{.2}^{\sqrt{2}-1}$

$=2^{\frac{1}{2}+\sqrt{2}-1}$

$=2^{\sqrt{2}-\frac{1}{2}}$