First slide

Introduction to limits

Question

$lim_{x \to -} {(\frac{x^{2} + 4 x - 3}{x^{2} - 2 x + 5})}^{x}$ is equal to

Moderate

A

$e^{6}$
B

$e^{- 6}$
C

$e^{2}$
D

$e^{4}$

Solution

We have,
$lim_{x \to \infty} {(\frac{x^{2} + 4 x - 3}{x^{2} - 2 x + 5})}^{x}$
$= lim_{x \to \infty} {(1 + \frac{6 x - 8}{x^{2} - 2 x + 5})}^{x} = e^{lim_{x \to \infty} \frac{6 x^{2} - 8 x}{x^{2} - 2 x + 5}} = e^{6}$

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