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

Question

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

How would you calculate it's limit? I thought using l'hopital rule, but it then becomes something nasty, as the differentiate of $(x + 1)^{\frac{1}{x}}$ isn't simple.

score 4 · Answer 1 · answered Jul 05 '15 at 20:22

4

Hint: $(1+x)^{1/x} = e^{(\ln(1+x))/x}.$ Use $\ln (1+h) = h-h^2/2 +O(h^3), e^h = 1 +h+O(h^2)$ for small $h.$

answered Jul 05 '15 at 20:22

zhw.

105,693

This is the way to go here - the way that I would approach it naturally. Inasmuch as you had already posted this, I tackled the problem "brute force" using L'Hospital's Rule, which shows the efficiency of asymptotics over The Old Hospital. +1 – Mark Viola Jul 05 '15 at 20:42

score 3 · Accepted Answer · answered Jul 05 '15 at 20:34

As @zhw already described, an efficient way to approach this is to use asymptotic analysis. So, here we will use instead the "brute force" approach via L'Hospital's Rule. To that end, we have

$$\begin{align} \lim_{x\to 0}\frac{(x+1)^{1/x}-e}{x}&=\lim_{x\to 0}(x+1)^{1/x}\left(\frac{x-(x+1)\log(x+1)}{x^2(x+1)}\right)\\\\ &=e\lim_{x\to 0}\left(\frac{x-(x+1)\log(x+1)}{x^2(x+1)}\right)\\\\ &=e\lim_{x\to 0}\left(\frac{-\log(x+1)}{3x^2+2x}\right)\\\\ &=e\lim_{x\to 0}\left(\frac{-\frac{1}{x+1}}{6x+2}\right)\\\\ &=-e/2 \end{align}$$

Bernard · Answer 3 · 2015-07-05T20:35:34.410

0

With Taylor's expansion at order $\mathit 1$: $$(1+x)^{\tfrac1x}=\mathrm e^\tfrac{\ln(1+x)}{x}=\mathrm e^{1-\tfrac{x}2+o(x)}=\mathrm e\Bigl(1-\frac x 2+o(x)\Bigr)$$ whence $$\frac{(1+x)^{\tfrac1x}-\mathrm e}x=\mathrm e\Bigl(-\frac12+o(1)\Bigr)\to-\frac{\mathrm e}2.$$

edited Jul 05 '15 at 20:35

answered Jul 05 '15 at 20:30

Bernard

175,478

That's what I wrote on paper, but could not retrieve where the $-$ sign came from :o( – Bernard Jul 05 '15 at 20:34
The sign error is due to the incorrect expansion of the $\log(1+x)$ which is $$\log(1+x)=x-\frac12x^2+O(x^3)$$. But how is this different from the approach by @zhw? – Mark Viola Jul 05 '15 at 20:36
Found it finally! Corrected the answer. Thank you for pointing the error. – Bernard Jul 05 '15 at 20:36
You're welcome. My pleasure. – Mark Viola Jul 05 '15 at 20:38
Dr MV: Thank you. You were posting while I was correcting :o) People are really fast these days! – Bernard Jul 05 '15 at 20:39
Usually faster than I ... ;-)) – Mark Viola Jul 05 '15 at 20:39

score 0 · Answer 4 · answered Jul 05 '15 at 22:41

I wouldn't say it is "nasty" to differentiate $f(x)=(1+x)^{\frac{1}{x}}$. We have:

$$ f'(x) = f(x)\cdot\frac{d}{dx}\log(f(x)) = (1+x)^{\frac{1}{x}}\cdot\frac{d}{dx}\frac{\log(1+x)}{x} $$ and since in a neighbourhood of the origin $\log(1+x)=x-\frac{x^2}{2}+o(x^2)$ we have: $$ \lim_{x\to 0^+} f'(x) = e\cdot\left(-\frac{1}{2}\right).$$

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

4 Answers4