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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?

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    $\begingroup$ WA probably only considers positive values of x (in which case the limit is indeed 1). $\endgroup$
    – Did
    Commented May 18, 2014 at 14:08
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    $\begingroup$ 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? $\endgroup$
    – Apurv
    Commented May 18, 2014 at 14:12
  • $\begingroup$ 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. $\endgroup$
    – Did
    Commented May 18, 2014 at 14:26

2 Answers 2

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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).

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  • $\begingroup$ 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. $\endgroup$
    – Did
    Commented May 18, 2014 at 14:24
  • $\begingroup$ ((Please do not erase your comments, especially when they are asking questions to which later comments answer.)) $\endgroup$
    – Did
    Commented May 18, 2014 at 14:37
  • $\begingroup$ 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) $\endgroup$
    – Apurv
    Commented May 18, 2014 at 14:48
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    $\begingroup$ 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 $x\to0$, $x\lt0$. $\endgroup$
    – Did
    Commented May 18, 2014 at 15:03
  • $\begingroup$ Thanks.. your comment answered my question.. $\endgroup$
    – Apurv
    Commented May 18, 2014 at 15:08
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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 $$\left(\frac{x}{e^{-x}+x-1}\right)^x=\exp\left(x\log \left(\frac{x}{e^{-x}+x-1}\right)\right)\sim_0\exp\left(x\log\left(\frac2x\right)\right)\xrightarrow{x\to0}e^0=1$$

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  • $\begingroup$ No it's $0$ and of course this limit exists when $\log x$ exists i.e. for $x>0$. $\endgroup$
    – user63181
    Commented May 18, 2014 at 14:17
  • $\begingroup$ 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? $\endgroup$
    – Apurv
    Commented May 18, 2014 at 14:19

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