How can I calculate $\lim\limits_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$?

Question

How can I calculate this limit?
$$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$$

I thought about L'Hospital because case of $\frac{0}{0}$, but I don't know how to contiune from this point..

Answer 1

Since $$\ln(x+1)= x-\frac{x^2}{2}+O(x^3)~~~and ~~~e^x= 1+x+\frac{x^2}{2}+O(x^3)$$ we get $$(1+x)^{\frac{1}{x}}= \exp\left(\frac{1}{x}\ln(1+x)\right) = \exp\left(\frac{1}{x}(x-\frac{x^2}{2} +O(x^3))\right) \\=\exp\left(1-\frac{x}{2} +O(x^2)\right) =e\exp\left(-\frac{x}{2}+O(x^2)\right) = e(1- \frac{x}{2}+O(x^2)) $$

Hence $$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x} = \lim_{x\rightarrow 0} \frac{e(1- \frac{x}{2}+O(x^2))-e}{x} =\lim_{x\rightarrow 0} -\frac{e}{2}+O(x) = \color{blue}{-\frac{e}{2}}$$

Answer 2

let $n = \frac 1x$

$\lim_\limits{n\to \infty} n((1+\frac 1n)^n - e)$

Binomial expansion:

$n(1 + 1 + \frac 12 (1-\frac 1n) + \frac 1{3!} (1-\frac 1n)(1-\frac 2{n})+\cdots +\frac {1}{n!} (1-\frac 1n)\cdots(1-\frac {n-1}{n}) - e)$

$e = 1 + 1 + \frac 1{2!} + \frac {1}{3!}\cdots$

$n( (-\frac 12 -\frac 1{3!} {3\choose 2} - \frac 1{4!} {4\choose 2} -\cdots - - \frac 1{n!} {n\choose 2})\frac 1{n} + o(\frac 1{n^2}))\\ -\frac 12(1+1+\frac 12 + \frac 1{3!}\cdots \frac {1}{n!}) + o(\frac 1n)\\ -\frac 12 e$

Answer 3

Using L'Hospital is one way to go here if you don't have knowledge of Taylor series yet.

The only tricky derivative is probably the derivative of $(1+x)^{\frac1x}$, which you can more easily calculate as

$$\left((1+x)^{\frac1x}\right)' = \left(e^{\frac1x\ln(1+x)}\right)' = e^{\frac1x\ln(1+x)} \cdot \left(\frac{1}{x(x+1)} - \frac{\ln(1+x)}{x^2}\right) = (1+x)^{\frac1x}\left(\frac{1}{x(x+1)} - \frac{\ln(1+x)}{x^2}\right)$$

Given that the first expression, $(1+x)^{\frac1x}$, has a limit of $e$, you only need to find the limit $$\lim_{x\to0}\left(\frac{1}{x(x+1)} - \frac{\ln(1+x)}{x^2}\right)$$ which is doable with a couple more applications of L'Hospital.

Answer 4

$$\begin{align}\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x} =& \lim_{x\rightarrow 0} \frac{\exp(1/x\ln(1+x))-e}{x}= e\lim_{x\rightarrow 0} \frac{\exp(1/x\ln(1+x)-1)-1}{x} &\\=& e\lim_{x\rightarrow 0} \frac{\exp(1/x\ln(1+x)-1)-1}{1/x\ln(1+x)-1 } \dfrac{1/x\ln(1+x)-1}{x}&\\=&e\lim_{x\rightarrow 0} \dfrac{\ln(1+x)-x}{x^2} &\\=& -\dfrac{e}{2}\end{align}$$

Limits used :

$\lim_{x \to 0} \dfrac{\ln(x+1)}{x} = 1$, $\lim_{h \to 0}\dfrac{e^h - 1}{h} = 1$ and $\lim_{x\to 0} \dfrac{\ln(x+1)-x}{x^2} = \dfrac{-1}{2}$

Are all limits solvable without L'Hôpital Rule or Series Expansion for the derivation of third limit without LH rule and series.

Answer 5

By the generalized binomial theorem,

$$(1+x)^{1/x}=1+\frac1xx+\frac1x\left(\frac1x-1\right)\frac{x^2}2+\frac1x\left(\frac1x-1\right)\left(\frac1x-2\right)\frac{x^3}{3!}+\cdots\\ =1+1+\frac{1-x}{2}+\frac{(1-x)(1-2x)}{3!}++\frac{(1-x)(1-2x)(1-3x)}{4!}\cdots$$

Clearly, the constant term is $$\sum_{k=0}^\infty\frac1{k!}=e.$$

Then the first degree coefficient is

$$-\frac12-\frac{1+2}{3!}-\frac{1+2+3}{4!}-\cdots=-\sum_{k=1}^\infty\frac{(k-1)k}2\frac1{k!}=-\frac e2.$$

Answer 6

By inequality

$$x-\frac{x^2}{2}\leq \log (1+x) \leq x-\frac{x^2}{2}+\frac{x^3}{3}$$

$$1-\frac{x}{2}\leq \frac{\log (1+x)}{x} \leq 1-\frac{x}{2}+\frac{x^2}{3}$$

we have that

$$e^{1-\frac{x}{2}}\leq (1+x)^\frac{1}{x}=e^{\frac{\log (1+x)}{x}} \leq e^{1-\frac{x}{2}+\frac{x^2}{3}} $$

thus

$$-\frac{e}2\le\frac{e^{1-\frac{x}{2}}-e}{x}\leq \frac{(1+x)^{\frac{1}{x}}-e}{x} \leq \frac{ e^{1-\frac{x}{2}+\frac{x^2}{3}}-e}{x}\to -\frac{e}2$$

indeed

$$ \frac{ e^{1-\frac{x}{2}+\frac{x^2}{3}}-e}{x} =e \cdot \frac{ e^{-\frac{x}{2}+\frac{x^2}{3}}-1}{ -\frac{x}{2}+\frac{x^2}{3} }\cdot\frac{-\frac{x}{2}+\frac{x^2}{3}}{x}\to e\cdot1\cdot -\frac12=-\frac{e}2$$

therefore for squeeze theorem

$$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}=-\frac{e}2$$

Answer 7

I just would like to add one idea to what's already been said. When using L'Hospital, never be mechanic. And I don't mean just to test that you have the right kind of indetermination before each time you apply the rule (checking hypotheses is a must before using any theorem/property). But it is also useful to watch in detail the function you are taking limits to and try to figure out where the difficulty is, so you can work apart with that specific expression. Taking derivatives of everything is not always of help; and it's better if you take the control of all those formulas instead the other way around.

For instance, the first time you use L'Hospital here you get the limit $$\lim_{x\to 0} \quad (1+x)^{\tfrac1x}\cdot \frac{x-(1+x)\log(1+x)}{x^2(1+x)}.$$ I put the first factor separate on purpose, because that one tends to $e$, while in the fraction both numerator and denominator go to $0$. Since the first factor bring no trouble, most of the times the better thing will be to leave it there and work separately with the quotient. Even more, the $(1+x)$ in the denominator is not what makes it $0$, so we could actually move it away too (and maybe cancel it with another $(1+x)$ from $(1+x)^{\tfrac1x}$).

So it's better to work with $$\lim_{x\to 0} \frac{x-(1+x)\log(1+x)}{x^2},$$ where we can use L'Hospital again: once you know where this tends to, you can put it together with the rest.

If you go on, you'll have again a similar situation, and so it will be important to think how you write things and what part you leave aside, too.

Of course, never forget to collect all the partial results at the end in order to give the solution of the original problem. (Yes, it's very easy to make such mistakes when you're nervous or short of time. =S)