Question

$\underset{x\to 0}{lim}\frac{x(e^{(\sqrt{1+x^2+x^4}-1)/x}-1)}{\sqrt{1+x^2+x^4}-1}$

• A. is equal to $\sqrt{e}$
• B. is equal to $1$
• C. is equal to $0$
• D. does not exist

Answer 1

The correct option is B is equal to $1$

$\underset{x\to 0}{lim}\frac{x(e^{(\sqrt{1+x^2+x^4}-1)/x}-1)}{\sqrt{1+x^2+x^4}-1}$
$=\underset{x\to 0}{lim}\frac{x(e^{\left[\frac{{\left(\sqrt{1+x^2+x^4}\right)}^2-1}{x(\sqrt{1+x^2+x^4}+1)}\right]}-1)\times (\sqrt{1+x^2+x^4}+1)}{(x^2+x^4)}$
$=\underset{x\to 0}{lim}\frac{x(e^{\left[\frac{1+x^2+x^4-1}{x(\sqrt{1+x^2+x^4}+1)}\right]}-1)\times (1+1)}{x(x+x^3)}$
$=\underset{x\to 0}{lim}\frac{(e^{\left[\frac{x(x+x^3)}{x(\sqrt{1+x^2+x^4}+1)}\right]}-1)\times 2}{(x+x^3)}$
$=\underset{x\to 0}{lim}\frac{e^{\left(\frac{x^3+x}{2}\right)}-1}{\left(\frac{x^3+x}{2}\right)\times 2}\times 2=1$