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I tried to evaluate $$\lim_{x\to0}\frac{(1+x)^{1/x}-e}x$$but all my efforts were in vain . L'Hospital rule becomes very complicated.

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Write, for $x > 0$, $$\begin{align*} \frac{(1+x)^{1/x}-e}x &= \frac{e}{x}\left( e^{\large\frac{1}{x}\ln(1+x)-1}-1 \right) \\ &=\frac{e}{x}\left( e^{\large\frac{x-\frac{x^2}{2}+o(x^2)}{x}-1}-1 \right) \\ &=\frac{e}{x}\left( e^{\large-\frac{x}{2}+o(x)}-1 \right) \\ &=\frac{e}{x}\left( 1-\frac{x}{2}+o(x)-1 \right) \\ &=e\left( -\frac{1}{2}+o(1) \right) \xrightarrow[x\to0^+]{}-\frac{e}{2} \end{align*} $$

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You want to find $L = \lim_{x\to0}\dfrac{(1+x)^{1/x}-e}x$.

Let $f(x) = (1+x)^{1/x}$. Since $\lim_{x\to0} f(x) = e$, by the definition of derivative, $L = f'(0)$.

Applying the chain rule $(f(g(x))' = g'(x)f'(g(x))$ in the form $(e^{g(x)})' = g'(x) e^{g(x)}$ very carefully,

$\begin{align} f'(x) &= ((1+x)^{1/x})'\\ &= (\exp((1/x)\ln(1+x))'\\ &= ((1/x)\ln(1+x))'(\exp((1/x)\ln(1+x))\quad\quad( (e^x)' = e^x)\\ &= ((1/x)/(1+x)-\ln(1+x)/(x^2))'(1+x)^{1/x}\quad(\text{Product rule})\\ &= \left(\dfrac{1}{x(1+x)}-\dfrac{\ln(1+x)}{x^2}\right)(1+x)^{1/x}\\ &= \left(\dfrac{x-(1+x)\ln(1+x)}{x^2(1+x)}\right)(1+x)^{1/x}\\ \end{align} $

Since $\lim_{x \to 0+} (1+x)^{1/x} = e$, we only need to evaluate the left-hand term as $x \to 0+$. For this, L'Hospital's rule rules.

$\begin{align} \lim_{x \to 0+} \dfrac{x-(1+x)\ln(1+x)}{x^2(1+x)} &=\lim_{x \to 0+}\dfrac{1-((1+x)/(1+x)+\ln(1+x)}{2x(1+x)+x^2}\\ &=\lim_{x \to 0+}\dfrac{-\ln(1+x)}{2x+x^2}\\ &=\lim_{x \to 0+}\dfrac{-1/(1+x)}{2+2x}\quad\quad\text{Applying Hoppy again}\\ &= -\dfrac{1}{2} \end{align} $

So $f'(0) = -\dfrac{e}{2}$.

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  • $\begingroup$ In fact, simply applying L'Hopital's Rule to the limit (as OP likely did) leads one to do exactly what is done here from computing $ \ f'(x) \ $ onward (that is, one is being asked to evaluate $ \ f'(0) \ ) \ . $ $\endgroup$ – colormegone Jul 19 '13 at 20:00
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By definition, $(1+x)^{1/x}=\exp(\frac1x\ln(1+x))$. Using Taylor series $$\ln(1+x)=x-\frac12x^2+O(x^3) $$ $$\exp(x)=1+x+\frac12x^2+O(x^3)$$ we find $$\begin{align}(1+x)^{1/x}-e&=e\cdot(\exp(\frac1x\ln(1+x)-1)-1)\\&=e\cdot(\exp(-1+1-\frac12x+O(x^2))-1) \\&= e\cdot(\exp(-\frac12x)-1+O(x^2))\end{align}$$ Then $ \frac{\exp(-\frac12x)-1}{x}$ tends (by definition) to the derivative of $x\mapsto \exp(-\frac12x)$ at $x=0$.

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