Limit as $x\to 0$ of $\frac{(1+x)^{1/x}-e}{x}$

Question

Find the following limit: $$\lim_{x\to 0}\frac{(1+x)^{1/x}-e}{x} \tag{1}$$

The following is my approach, although is full of incorrect assumptions (statements etc). $$f(x)= \lim_{h\to 0}(1+x+h)^{1/(x+h)}\\$$

From here we can say $f(0) = e$.

$$\ln(f(x)) = \lim_{h\to 0}\frac{1}{x+h}\ln(1+x+h)$$

Near $x=0$, we can use the series of logarithm: $$\ln(f(x)) = \lim_{h\to 0}\frac{1}{x+h}\left(x+h-\frac{(x+h)^2}{2}...\right)$$

Differentiating we get:

$$\frac{f'(x)}{f(x)} = \lim_{h\to 0}-\frac{1}{2} +\frac{x+h}{3} ... \tag{2}$$

Now we note that $(1)$ is actually $f'(0)$ (?) and so from $(2)$ we get:

$$f'(0) =\frac{-1}{2} f(0) = \frac{-e}{2}$$

While the answer is seemingly correct, the method is absolutely not. It looks like this is fluke than anything else.

I also tried computing taylor series of $(1+x)^{1/x}$ near $x=0$ but couldn't do it.

Answer 1

You are making things complicated by bringing in the $h$. One has $$\ln[(1+x)^{1/x}]=1-\frac x2+O(x^2)$$ so taking exponentials gives $$(1+x)^{1/x}=e\exp(-x/2+O(x^2))=e(1-x/2+O(x^2)).$$ Then $$\frac{(1+x)^{1/x}-e}{x}=-\frac e2+O(x)$$ etc. This is really the same sort of manipulation as you were doing...

Answer 2

Hint

Consider $$y=(1+x)^{1/x}\implies \log(y)=\frac 1x \log(1+x)$$ Now, using Taylor $$\log(y)=\frac 1x\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$ Now $$y=e^{\log(y)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$ Just continue.

Edit

For the limit itself, the development to $O\left(x^2\right)$ was enough. Doing it to $O\left(x^3\right)$ allows to find the limit and also how it is approached.

Answer 3

Why don't you just take $e$ as a factor and transform the given expression into $$e\cdot\dfrac{\exp\left(\dfrac{\log(1+x)}{x}-1\right) - 1} {\dfrac{\log(1+x)}{x}-1}\cdot\frac{\log(1+x)-x}{x^2} $$ The middle factor tends to $1$ because argument of $\exp$ tends to $0$. The last factor tends to $-1/2$ via Taylor series or L'Hospital's Rule so that the final answer is $-e/2$.

Answer 4

Using LH rule,

$$ \begin{align} & \lim_{x \to 0} \dfrac{(1 + x)^{1/x} - e}x \\ &= \lim_{x \to 0} \frac{(1+x)^{1/x} \left(\frac{-\ln(x + 1)}{x^2}+\frac1{x(x + 1)}\right) - 0}1\\ &= \lim_{x \to 0} (1 + x)^{1/x}\left( \color{brown}{\frac{x - \ln(x + 1)}{x^2}} - \frac1{x + 1}\right)\\ &= e \left(\frac12 - 1\right) \\ &= \frac{-e}2 \end{align} $$

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$$ \lim_{x \to 0} \color{brown}{\frac{x - \ln(x + 1)}{x^2}} = \lim_{x \to 0} \frac{1 - \frac 1{1+x}}{2x} = \frac 12. $$

Answer 5

Binomial expansion way: $$(1+x)^{\frac1x}=\sum_{k=0}^\infty{\frac1x\choose k}x^k=2+\sum_{k=2}^\infty\frac{(1-x)(1-2x)...(1-(k-1)x)}{k!}\tag1$$ $$e=2+\sum_{k=2}^\infty\frac1{k!}\tag2$$ On the other hand, for $k\geq2$, $$\lim_{x\to 0} \frac{(1-x)(1-2x)...(1-(k-1)x)-1}{k!}=-\frac{1}{2(k-2)!}$$ From $(1)$, $(2)$ and $(3)$ the limit is calculated to be $$\lim_{x\to0}\frac{(1+x)^{\frac1x}-e}{x}=\sum_{k=2}^{\infty}-\frac{1}{2(k-2)!}=-\frac e2.$$

Answer 6

The binomial expansion of $(1+x)^\frac 1x$ using the binomial theorem.

$(1+x)^\frac 1x = 1 + (\frac 1x)x + (\frac 12)(\frac 1x)(\frac 1x -1) x^2 + (\frac 1{3!})(\frac 1x)(\frac 1x -1)(\frac 1x - 2) x^3 + \cdots\\ 1 + 1 + (\frac 12)(1 -x) + (\frac 1{3!})(1 -x)(1-2x) + (\frac 1{4!})(1 -x)(1-2x)(1-3x)+\cdots$

The constant term of this series is $1+ 1+\frac 12 + \frac {1}{3!} + \frac {1}{4!} + \cdots = e$

What is the coefficient of the $x$ term?

$-\frac 1{2!} x - \frac {1}{3!} (1+2) x - \frac {1}{4!} (1+2+3) x - \frac {1}{5!} (1+2+3+4)x - \cdots$

$\sum_\limits{k=2}^\infty -\frac {1}{k!} \frac {(k)(k-1)}{2} = -\sum_\limits{k=2}^\infty \frac {1}{2(k-2)!} = -\frac {e}{2}$

The $x^2$ and higher terms we don't care about.