Find the limit: $$\lim_\limits{x\to 0}{\frac{\left(1+x\right)^{1/x}-e}{x}}$$
I have no idea what to do, but I thought that this is the limit of the derivative of $f(x)=\left(1+x\right)^{1/x}$, as $x$ tends to 0. Any help?
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You are right. Note that $(1+x)^{1/x}=e^{(1/x)\ln(1+x)}$. – André Nicolas Mar 10 '16 at 18:18
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@AndréNicolas But I cannot use De L' Hospital, because the function of the numerator is not differentiable at 0. Also, the limit of the derivative is more complex than the original one... – Mar 10 '16 at 18:24
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@Dr.MV Obviously it does not exist at 0... – Mar 10 '16 at 18:25
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Yes, it actually does exist. – Mark Viola Mar 10 '16 at 18:29
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Without Taylor please, because I haven't been taught that yet... – Mar 10 '16 at 18:34
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I prefer series, but L'Hospital's Rule works quite well here. – André Nicolas Mar 10 '16 at 18:35
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1You can also differentiate $(1+x)^{1/x}$ (for $x\neq 0$ and take the right and left limits at $0$. – Clement C. Mar 10 '16 at 18:37
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See related http://math.stackexchange.com/a/1314198/72031 – Paramanand Singh Mar 11 '16 at 02:08
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See also: How to solve this limit: $\lim\limits_{x\to0}\frac{(1+x)^{1/x}-e}x$? – Martin Sleziak Dec 06 '19 at 07:01
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Indeed, it is: define the function $f$ on $\mathbb{R}$ by $$ f(x) = \begin{cases}(1+x)^{1/x} &\text{ if } x \neq 0 \\ e &\text{ if } x=0\end{cases}. $$ Then you can check that $f$ is continuous on $\mathbb{R}$, and what you want is exactly the definition of $f^\prime(0)$ (which is not guaranteed to exist: the question asks to both show its existence, and to find its value.)
To find the limit (show it exists and what its value is), I would advocate a systematic, general method for these sort of problems, namely Taylor expansions. It requires a bit of background, but once assimilated makes these limits much easier to compute:
We will need the following two expansions, where $o(\cdot)$ is the Landau notation ("little-o"), here taken when $u\to 0$: $$ \begin{align} \ln(1+u) &= u-\frac{u^2}{2} + o(u^2) \\ e^u &= 1+u + o(u) \end{align} $$ which you can interpret as "approximations of the function on the LHS around 0 by polynomials" (here, respectively degree-2 polynomial and affine function). The $o(u^2)$ 'means,' in a very formal and rigorous sense, "goes to zero much faster than $u^2$, when $u\to 0$."
Then, for any $x\neq 0$ $$ (1+x)^{1/x} = e^{\frac{1}{x}\ln(1+x)} $$ and since $\ln(1+u) = u - \frac{u^2}{2} + o(u^2)$ when $u\to 0$, you get $$ (1+x)^{1/x} = e^{\frac{1}{x}(x-\frac{x^2}{2}+o(x^2))} = e^{1-\frac{x}{2}+o(x)} = e\cdot e^{-\frac{x}{2}+o(x)}. $$ Now, we use that when $u\to 0$, we have $e^u = 1+u+o(u)$: since $-\frac{x}{2}+o(x) \to 0$ when $x\to 0$, we obtain $$ (1+x)^{1/x} = e\cdot e^{-\frac{x}{2}+o(x)} = e\cdot\left(1-\frac{x}{2}+o(x)\right) = e - \frac{ex}{2} + o(x). $$
It only remains to put it all together: $$ \frac{(1+x)^{1/x}-e}{x} = \frac{e - \frac{ex}{2} + o(x)-e}{x} = - \frac{e}{2} + o(1) \xrightarrow[x\to 0]{} - \frac{e}{2}. $$
Clement C.
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You can also do it with the "Theorem of the limit of the derivative." Here, $f$ is continuous on $\mathbb{R}$, and differentiable on $\mathbb{R}^\ast$. If you can show that $f^\prime$ has a limit voth at $0^+$ and $0^-$, and that these two limits are equal, then you are done: $f$ will be differentiable at $0$, and $f^prime(0)$ will be this limit. So you need to differentiate $f$ for $x\neq 0$, and take the two limits. – Clement C. Mar 10 '16 at 18:36
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How would I take the limits, since after differentiation those appear more complex than the original limit? – Mar 10 '16 at 18:38
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I didn't go through the computations, TBH -- you ask without Taylor, this is the thing that comes to mind: it might itself not be easy if you don't want to use Taylor nor L'Hopital. – Clement C. Mar 10 '16 at 18:40
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I didn't say that we needn't use DLH. We may use that, since we have been taught that. – Mar 10 '16 at 18:41
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Then, you may try to use it there... differentiate $f$, and apply L'Hopital's rule (maybe several times) to find the limit of this derivative at $0$. Did you get stuck somewhere? – Clement C. Mar 10 '16 at 18:45
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Yes, at this one here: $$\lim_\limits{x\to 0}{\frac{x-(1+x)\ln (1+x)}{x^2}}$$
Isn't there any smart way to do that, apart from doing two tedious DLH?
– Mar 10 '16 at 18:57 -
1Yes, see André Nicolas's answer (which relies on the known limit of $(1+x)^{1/x}$ to remove as much of possible of the tediousness): http://math.stackexchange.com/a/1691968/75808 – Clement C. Mar 10 '16 at 19:03
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If Taylor series (the "right" approach) are not yet available, let's use L'Hospital's Rule.
When we differentiate the top we get $$(1+x)^{1/x}\left(\frac{x/(1+x)-\ln(1+x)}{x^2}\right).$$ The front part safely has limit $e$, so we only need to find $$\lim_{x\to 0}\frac{x/(1+x)-\ln(1+x)}{x^2}.$$ One round of L'Hospital's Rule gets us to $$\lim_{x\to 0}\frac{1/(1+x)^2-1/(1+x)}{2x}.$$ Now a little algebra finishes things. Or if one likes L'Hospital's Rule, do it again.
André Nicolas
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