Evaluate $\lim_{x \to 0} \frac{e - (1 + x)^{\frac{1}{x}}}{\tan x} $
Answer:
$\lim_{x \to 0} \frac{e - (1 + x)^{\frac{1}{x}}}{\tan x} $
$= \lim_{x \to 0} \cos x \cdot \frac{e - (1 + x)^{\frac{1}{x}}}{\sin x} $
$= \lim_{x \to 0} \cos x \cdot\ \lim_{x \to 0} \frac{e - (1 + x)^{\frac{1}{x}}}{\sin x} $
$= \lim_{x \to 0} \frac{e - (1 + x)^{\frac{1}{x}}}{\sin x} $
$= \lim_{x \to 0} \frac{(1+x)^{\left(\frac{1}{x} - 1\right)}((1+x) \ln(1+x) -x))}{x^2\cos x} $ (LHopital Rule)
$= \lim_{x \to 0} \frac{(1+x)^{\left(\frac{1}{x} - 1\right)}((1+x) \ln(1+x) - x))}{x^2} $
$= \lim_{x \to 0} (1+x)^{\left(\frac{1}{x} - 1\right)} \cdot \lim_{x \to 0} \frac{ (1+x) \ln(1+x) - x}{x^2} $
$ \lim_{x \to 0} (1+x)^{\left(\frac{1}{x} - 1\right)}$
$= \lim_{x \to 0} (1+x)^{\frac{1}{x}} (1 + x)^{- 1}$
$= \lim_{x \to 0} (1+x)^{\frac{1}{x}} \lim_{x \to 0} (1 + x)^{- 1}$
$= e \cdot 1 = e $
and,
$\lim_{x \to 0} \frac{(1+x) \ln(1+x) - x}{x^2} $
$= \lim_{x \to 0} \frac{\ln(x+1)}{2x} $ (LHopital Rule)
$= \lim_{x \to 0} \frac{1}{2(x+1)} $ (LHopital Rule)
$= \lim_{x \to 0} \frac{1}{2} $
So,
$= \lim_{x \to 0} (1+x)^{(\frac{1}{x} - 1)} \cdot \lim_{x \to 0} \frac{ (1+x) \ln(1+x) - x}{x^2} = \frac{e}{2}$
