Evaluating $ \lim_{x \to 0} \frac{e - (1 + 2x)^{1/2x}}{x} $ without using any expansion series

Question

The following is a question from the Joint Entrance Examination (Main) from the 09 April 2024 evening shift:

$$ \lim_{x \to 0} \frac{e - (1 + 2x)^{1/2x}}{x} $$ is equal to:

(A) $0$

(B) $\frac{-2}{e}$

(C) $e$

(D) $e - e^2$

I rearranged the problem to this:

$$ \lim_{x \to 0} \frac{e - e^{\log(1 + 2x)/(2x)}}{x} $$

And tried to evaluate the limit, but I am stuck. I don't want the solution, a pointer towards the right way is sufficient. Please try not to evaluate using any expansion series, unless it's absolutely impossible to solve the limit without using it. Any other methods are welcome.

Can you accept to set $\frac1{2x}=n$ and develop a few terms? — , Sep 11 '24 at 15:28
L'Hopital's rule will work. Also remember that $\lim_{z \to 0} (1+z)^{1/z} = e$, which might be more familiar in the form $\lim_{x \to \infty} (1+1/x)^x = e$. — Michael Lugo, Sep 11 '24 at 15:32
Can anyone please tell me exactly what I can do to reopen my question, and if I can do anything at all? — archie, Sep 24 '24 at 14:29

score 4 · Accepted Answer · answered Sep 11 '24 at 16:54

Your start is good, then we can proceed as follows

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

and then using standard limit $\frac{e^x-1}x\to 1$ and $\frac{\log(1+x)-x}{x^2}\to -\frac12$ (see here).

score 2 · Answer 2 · 2024-09-11T16:40:52.813

L'Hospital is not very easy in this case. For convenience, we will first transform to the limit of

$$2\frac{e-(1+x)^{1/x}}{x}$$ and keep the $2$ aside.

Now we differentiate the numerator (the denominator vanishes), giving

$$\left(\frac{\log(x+1)}{x^2}-\frac1{x(x+1)}\right)(x+1)^{1/x}.$$

This is again indeterminate, but we may evaluate the right factor away, giving $e$. Then we rewrite the expression as

$$\frac{\log(x+1)-\dfrac x{x+1}}{x^2}.$$

With one more application of L'Hospital,

$$\frac{\dfrac1{x+1}-\dfrac1{(x+1)^2}}{2x}$$ which tends to $\dfrac12$.

The final answer is $e$.

With Taylor anyway:

$$\frac{\log(1+2x)}{2x}=\frac{2x-\dfrac{(2x)^2}2+\cdots}{2x}=1-x+\cdots$$

Taking the antilogarithm,

$$(1+x)^{1/x}=e\cdot e^{-x+\cdots}=e(1-x+\cdots)$$

and the limit is that of

$$\frac{e-e+ex+\cdots}x.$$

There is no need to compute higher order terms.

score 2 · Answer 3 · answered Sep 11 '24 at 16:57

First of all, substitute $2x\mapsto x$, so the limit becomes: $$L=\frac12\lim_{x\to 0} \frac{e-(1+x)^{1/x}}x=\frac e2\lim_{x\to 0}\frac 1x\left( 1- \exp\left(\frac1x\log(1+x)-1\right)\right) $$ Now note that $\frac1x\log(1+x)-1\to 0$ and use the fact that

$$\lim_{x\to 0} \frac{e^{f(x)}-1}{f(x)}=1 \quad \text{if}\quad f(x)\to 0$$ Letting $f(x)=\frac1x\log(1+x)-1$ you have: $$L=-\frac e2\lim_{x\to 0}\frac1x \frac{e^{f(x)}-1}xf(x)=-\frac e2\lim_{x\to 0}\frac{\log(1+x)-x}{x^2}=e$$ To evaluate the last limit without series expansion, see for example this question.

Evaluating $ \lim_{x \to 0} \frac{e - (1 + 2x)^{1/2x}}{x} $ without using any expansion series

3 Answers3