For x∈R,limx→∞ x−3x+2x=

A
$e$
B
$e^{- 1}$
C
$e^{- 5}$
D
$e^{5}$

Solution:

We have,

$lim_{x \to \infty} {(\frac{x - 3}{x + 2})}^{x} = lim_{x \to \infty} {(1 + \frac{- 5}{x + 2})}^{x}$

$\Rightarrow lim_{x \to \infty} {(\frac{x - 3}{x + 2})}^{x} = e^{lim_{x \to \infty} - 5 x} x + 2 = e^{- 5} [lim_{x \to \infty} \frac{- 5}{1 + \frac{2}{x}} = - 5]$

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