Find the value of the limit $$\lim_{x\to0}\left(\frac{e-(1+x)^\frac{1}{x}}{x}\right).$$
I tried to apply the standard limit $$\lim_{x\to0}(1+x)^\frac{1}{x} = e$$ and L'Hospital's Theorem individually, but that didn't help me.
Any help will be appreciated.
Best Answer
Using the standard Taylor expansions (to low order) $\ln(1+u)=u-\frac{u^2}{2}+o(u^2)$ and $e^u=1+u+o(u)$ when $u\to0$.
Rewriting $$\begin{align} (1+x)^{\frac{1}{x}} &= \exp\left( \frac{1}{x}\ln(1+x)\right) = \exp\left( \frac{1}{x}(x-\frac{x^2}{2} + o(x^2))\right) = \exp\left( 1-\frac{x}{2} + o(x)\right) \\&= e\cdot \exp\left( -\frac{x}{2} + o(x)\right) = e\cdot \left( 1-\frac{x}{2} + o(x)\right) = e-e\frac{x}{2} + o(x) \end{align}$$ we get that $$ \frac{e-(1+x)^{\frac{1}{x}}}{x} = \frac{e\frac{x}{2} + o(x)}{x} =\frac{e}{2} + o(1)\xrightarrow[x\to0]{} \boxed{\frac{e}{2}}. $$
Note that we expanded $\ln(1+x)$ to order $x^2$, since we can "guess" the first order will get cancelled eventually by the $-e$ in the denominator. (Doing only the expansion to first order will only, basically, yield the limit $(1+x)^{1/x}\xrightarrow[x\to0]{}e$, and so we know we need better precision.)
In the second step, factoring the $e$ out of the product allows us to get $e^{-x/2+o(x)}$ instead of $e^{1-x/2+o(x)}$, which is required in order to use the expansion of $e^u$ (since this expansion holds when $u\to 0$, and while $\frac{x}{2}\to 0$, this is not the case for $1-\frac{x}{2}$.)