Question

Evaluate:

$\underset{x\to 0}{lim}{[1+x\ln (1+b^2)]}^{\frac{1}{x}}$ is equal to

• A. $1+b^2$
• B. $1-b^2$
• C. $e^{1-b^2}$
• D. $e^{1+b^2}$

Answer 1

The correct option is A $1+b^2$

LHS

$\Rightarrow \underset{x\to 0}{lim}{e}^{\ln {[1+x\ln (l+b^2)]}^{\frac{1}{x}}}$

$\underset{x\to 0}{lim}{e}^{\frac{1}{x}\ln [1+xln(l+b^2)]}$

$\underset{x\to 0}{lim}\frac{\ln [1+xln(l+b^2)]}{x}$

Now it form's $\frac{0}{0}$

Apply L'Hopital Rule

Differentiate numerator and denominator w.r.t $x$

$\Rightarrow \underset{x\to 0}{lim}\frac{\ln (l+b^2)}{1+x\ln (l+b^2)}$

$=e^{\ln (l+b^2)}$

$=1+b^2$.