A limit problem by Sladjan

Evaluate $ \lim_{x \to 0} \frac{(1 + x)^{\frac{1}{x}} - e}{x} $

Answer:

Let $ L = \lim_{x \to 0} \frac{(1 + x)^{\frac{1}{x}} - e}{x} $

Since $ \lim_{x \to 0} (x + x)^{\frac{1}{x}} - e = 0 $

and

$ \lim_{x \to 0} x = 0 $

So, we can apply LHopital Rule and we have:

$L = \lim_{x \to 0} \frac{\frac{d}{dx}\left( (1 + x)^{\frac{1}{x}} - e\right) }{\frac{d}{dx}(x)} $

$= \lim_{x \to 0} \frac{(1 + x)^{\frac{1}{x}} \left(\frac{1}{x(x+1)} - \frac{1}{x^2}\ln(1 + x) \right)  }{1} $

$= \lim_{x \to 0} (1 + x)^{\frac{1}{x}} \left(\frac{1}{x(x+1)} - \frac{1}{x^2}\ln(1 + x) \right) $

$= \lim_{x \to 0} (1 + x)^{\frac{1}{x}} \lim_{x \to 0} \left(\frac{1}{x(x+1)} - \frac{1}{x^2}\ln(1 + x) \right) $

$= \lim_{x \to 0} (1 + x)^{\frac{1}{x}} \lim_{x \to 0} \frac{x - (1+x)\ln(1+x)}{x^2(1 + x)}  $

Apply LHopital Rule to second limit and get

$= (e )* \lim_{x \to 0} \frac {1 - 1 - ln(1 + x)}{2x+3x^2} $

Apply LHopital Rule again to second limit and get

$= (e )* \lim_{x \to 0} \frac {\frac{-1}{1 + x}}{2+6x} $

$= \frac{-e}{2} $