Question

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

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

Answer 1

Answer: (D)

$\underset{x\to 0}{\lim }\frac{x\left(e^{\left(\sqrt{1+x^2+x^4}-1\right)/x}-1\right)}{\sqrt{1+x^2+x^4}-1}$
$\because \underset{x\to 0}{\lim }\frac{\sqrt{1+x^2+x^4}-1}{x}\left(\frac{0}{0}\text{ from }\right)$
$\underset{x\to 0}{\lim }\frac{\left(1+x^2+x^4\right)-1}{x(\sqrt{1+x^2+x^4}+1}$
$\underset{x\to 0}{\lim }\frac{x\left(1+x^2\right)}{\left(\sqrt{1+x^2+x^4}+1\right)}=0$
So $\underset{x\to 0}{\lim }\frac{x\left(e^{\left(\frac{\sqrt{1+x^2+x^4}-1}{x}\right)}-1\right)}{\sqrt{1+x^2+x^4}-1}\left(\frac{0}{0}\text{ from }\right)$
$\underset{x\to 0}{\lim }\frac{e^{\frac{\sqrt{1+x^2+x^4-1}}{x}}-1}{\left(\frac{\sqrt{1+x^2+x^4}-1}{x}\right)}=1$