Question

$\underset{x\to \infty }{lim}{\left(\frac{2x^2+3}{2x^2+5}\right)}^{8^{x^2}+3}$ is equal to.

• A. $0$
• B. $1$
• C. $e^8$
• D. $e^{-8}$

Answer 1

The correct option is A $0$

Let the given limit be $L$

We have $L=\underset{x\to \infty }{lim}{\left(\frac{2x^2+3}{2x^2+5}\right)}^{8^{x^2}+3}$

Observe that given limit is in the form of $1^{\infty }$

$\Rightarrow L=e^{{lim}_{x\to \infty }(8^{x^2}+3)(\frac{2x^2+3}{2x^2+5}-1)}=e^{{lim}_{x\to \infty }(8^{x^2}+3)\left(\frac{-2}{2x^2+5}\right)}$

$\Rightarrow L=e^{-2\left({lim}_{x\to \infty }\left(\frac{8^{x^2}+3}{2x^2+5}\right)\right)}$

Now observe that as $x\to \infty$ , the value of $L$ tends to

$e^{-\infty }=0$

Therefore the correct option is $A$