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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.

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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 $1-\frac{x}{2}$.)

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Hint. Expand the function $f(x):=(1+x)^{\frac{1}{x}}=\exp\left(\frac{\ln(1+x)}{x}\right)$ at $0$: $$f(x)=\exp\left(\frac{x-\frac{x^2}{2}+o(x^2)}{x}\right)=\exp\left(1-\frac{x}{2}+o(x)\right)=e\cdot \exp\left(-\frac{x}{2}+o(x)\right).$$ Can you take it from here? At the end you should find that the limit is $e/2$.

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Put $$f (x)=(1+x)^\frac 1x $$ for $x\ne 0$ and $f (0)=e $.

then your limit is $$-\lim_0\frac {f (x)-f (0)}{x-0} $$ or $$-\lim_0 f'(x ).$$ with

$f'(x)=f (x)(-\frac {1}{x^2}\ln (1+x)+\frac {1}{x (1+x)}) $

$=f (x)(\frac {-x+x^2/2 (1+\epsilon (x))}{x^2}+\frac {1}{x}-\frac {1}{1+x}) $

from this, the limit is $$-e(\frac {1}{2}-1)=\frac {e}{2} $$

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  • $\begingroup$ Not that I want to nitpick, but you actually have to first show that $f$ is differentiable at $0$, don't you? (the step $f'$ exists for every $x>0$, and has a limit at $0$, implies that $f'$ is continuously differentiable at $0$ might not be obvious) $\endgroup$
    – Clement C.
    Jun 5 '17 at 12:24
  • $\begingroup$ @ClementC. My method is based on MVT. $\endgroup$ Jun 5 '17 at 12:35
  • $\begingroup$ it is basically the application of the L'opital's rule, which the asker could not carry out $\endgroup$
    – farruhota
    Jun 5 '17 at 12:36
  • $\begingroup$ Not saying it isn't, or trying to infer what the underlying of your proof are. My point is that your answer would be even better if you mentioned why this step can be made, even concisely. (Since, as it stands, the gap lies in the "from this" step -- the OP was confused about another point, and it's fair to assume they might be confused by this one.) $\endgroup$
    – Clement C.
    Jun 5 '17 at 12:36
  • $\begingroup$ @FarrukhAtaev No not l'Hopital, simple MVT . $\endgroup$ Jun 5 '17 at 12:38

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