# What is $\lim\limits_{x\to 0}\left(\dfrac{x}{e^{-x}+x-1}\right)^x$

What is

$$\lim_{x\to 0}\left(\frac{x}{e^{-x}+x-1}\right)^x$$

Using the expansion of $e^x$, I get that the function

$$y=\left(\frac{x}{e^{-x}+x-1}\right)^x$$

is not defined for negative numbers.

Hence the limit at $0^{-}$ must not exist.$\implies$The limit at $0$ does not exist.

However WA says that it should be $1$. :(

Am I wrong?

• WA probably only considers positive values of x (in which case the limit is indeed 1).
– Did
Commented May 18, 2014 at 14:08
• But for the limit to be defined, the LHL and RHL must both exist and be equal. If WA doesn't consider that, then is WA wrong? Commented May 18, 2014 at 14:12
• Actually I withdraw my first comment. Look at the diagram on the WA page: WA considers bilateral limits and interprets powers of negative real numbers as complex logarithms, see my answer.
– Did
Commented May 18, 2014 at 14:26

WA interprets the number $$u(x)=\left(\frac{x}{\mathrm e^{-x}+x-1}\right)^x$$ when $x\gt0$ as $$u(x)=\exp\left(x\log\left(\frac{x}{\mathrm e^{-x}+x-1}\right)\right),$$ and when $x\lt0$ as $$u(x)=\exp\left(x\log\left(\frac{-x}{\mathrm e^{-x}+x-1}\right)+\mathrm i\pi x\right).$$ Then both limits are indeed $1$ (as one sees when one looks closely at the plot on the WA page).

• Nobody is talking abount complex nonreal arguments $x$ here, only the function has limits when $x\to0$ with $x$ positive real and when $x\to0$ with $x$ negative real, and these two limits coincide.
– Did
Commented May 18, 2014 at 14:24
– Did
Commented May 18, 2014 at 14:37
• Pardon me, but I still don't understand why the limit for a function exists at a point when you can approach the point from only one side.(the function is real valued) Commented May 18, 2014 at 14:48
• Did you read my answer? Either one defines a real valued function on $x$ real, $x\gt0$ only, then the limit when $x\to0$ can only be (obviously) a limit from the right. Or, one defines a complex valued function on $x$ real, $x\gt0$ and $x\lt0$, then one obtains a function which happens to be real valued on $x\gt0$ and to be complex-valued on $x\lt0$ and which happens to have the both-sided limit $1$ at $0$, that is, when $x\to0$, $x\gt0$, and when $x\to0$, $x\lt0$.
– Did
Commented May 18, 2014 at 15:03
• Thanks.. your comment answered my question.. Commented May 18, 2014 at 15:08

We have using the Taylor series

$$e^{-x}+x-1\sim_0\frac{x^2}{2}$$ hence $$\frac{x}{e^{-x}+x-1}\sim_0\frac2x$$ and then $$\left(\frac{x}{e^{-x}+x-1}\right)^x=\exp\left(x\log \left(\frac{x}{e^{-x}+x-1}\right)\right)\sim_0\exp\left(x\log\left(\frac2x\right)\right)\xrightarrow{x\to0}e^0=1$$

• No it's $0$ and of course this limit exists when $\log x$ exists i.e. for $x>0$.
– user63181
Commented May 18, 2014 at 14:17
• then you can't call it a limit at $0$. What I am asking is that does the limit at $0$ exists? (The RHL may exist). Is it a matter of using language? Commented May 18, 2014 at 14:19