Question

The value of $\underset{x\to \infty }{lim}{\left[\frac{2x^2+3}{2x^2+5}\right]}^{8x^2+3}$ is equal to

• A. $e^{-8}$
• B. $e^8$
• C. $e^3$
• D. None

Answer 1

Answer: (A)

$e^{-8}$

$\underset{x\to \infty }{lim}{\left[\frac{2x^2+3}{2x^2+5}\right]}^{8x^2+3}$

=$1^{\infty }$ - Indeterminate form (Putting x equals $\infty$ )

We use following formula to solve problems of form $1^{\infty }$ ,

$\underset{x\to \infty }{lim}f(x{)}^{g\left(x\right)}$ = $e^{{lim}_{x\to \infty }\left[f\right(x)-1]g\left(x\right)}$

Thus, $\underset{x\to \infty }{lim}{\left[\frac{2x^2+3}{2x^2+5}\right]}^{8x^2+3}$

= $e^{\underset{x\to \infty }{lim}\left[\right(\frac{2x^2+3}{2x^2+5}-1\left)\right](8x^2+3)}$

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

=$e^{\underset{x\to \infty }{lim}-2\left(\frac{8+3/x^2}{2+5/x^2}\right)}$ (Dividing numerator and denominator both by$x^2$)

=$e^{-8}$ (Putting x equal to $\infty$)