I have come across a problem which requires solving the following limit without L'Hôpital rule:

$$\lim_{x\to\infty} x^2\cdot(e^\frac{1}{x-1}-e^\frac{1}{x})$$

It is obvious from the graphic plot (or using L'Hôpital rule) that the limit is 1. I have tried a few algebraic manipulations and changes of variable without success. Another approach that I tried was to "sandwich" this limit between two other different limits, one strictly greater and one strictly lesser than this one, that would be easier to work with than the difference of exponentials present in this one. As of now I haven't had any success, how would one go about solving it?


5 Answers 5


Let $x=\frac 1y$ to make $$ x^2\,\left(e^\frac{1}{x-1}-e^\frac{1}{x}\right)=\frac{e^{\frac{y}{1-y}}-e^y}{y^2}$$

Now, since $y$ is small $$\frac{y}{1-y}=y+y^2+O\left(y^3\right)$$ $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ $$e^{\frac{y}{1-y}}=1+y+\frac{3 y^2}{2}+O\left(y^3\right)$$ $$e^{\frac{y}{1-y}}-e^y=y^2+O\left(y^3\right)$$


If you are patient and then use more terms, you would find the nice $$\frac{e^{\frac{y}{1-y}}-e^y}{y^2}=1+2 y+3 y^2+O\left(y^3\right)$$ For $x=100$, $y=\frac 1 {100}$, this would give exactly $1.0203$ while the exact result would be $\approx 1.020304224$


$$\lim_{x\rightarrow\infty}x^2\left(e^{\frac{1}{x-1}}-e^{\frac{1}{x}}\right)=\lim_{x\rightarrow0}\frac{e^{\frac{x}{1-x}}-e^x}{x^2}=\lim_{x\rightarrow0}e^x\lim_{x\rightarrow0}\frac{e^{\frac{x}{1-x}-x}-1}{x^2}=$$ $$=\lim_{x\rightarrow0}\left(\frac{e^{\frac{x^2}{1-x}}-1}{\frac{x^2}{1-x}}\cdot\frac{1}{1-x}\right)=1$$

  • $\begingroup$ There must be some error, since the limit should be 1. $\endgroup$
    – Alex Jones
    Mar 2, 2018 at 13:23
  • 1
    $\begingroup$ @AlexanderJ93 Thank you! I fixed. See now. $\endgroup$ Mar 2, 2018 at 13:25
  • $\begingroup$ Isn't the stacked fraction on the left in the last bracket essentially "0/0"? $\endgroup$ Mar 2, 2018 at 16:28
  • 2
    $\begingroup$ @Max Freiburghaus I used $\lim\limits_{x\rightarrow0}\frac{e^x-1}{x}=1,$ which is $(e^x)'_{x=0}$ by the definition of the derivative. $\endgroup$ Mar 2, 2018 at 16:32

By first order expansion for exponential

  • $e^\frac{1}{x-1}\sim 1+\frac{1}{x-1}$
  • $e^\frac{1}{x}\sim 1+\frac{1}{x}$


$$x^2\cdot(e^\frac{1}{x-1}-e^\frac{1}{x})\sim x^2\cdot \left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{x^2}{x(x-1)}\to 1$$


Let $f(x)=e^\frac{1}{x}$. This is concave for large $x$, meaning that a difference quotient between two points is bounded between $f'$ evaluated at either point: $$f'(x-1)\leq\frac{f(x)-f(x-1)}{1}\leq f'(x) $$ (all terms above being negative)

With $f'(x)=-e^\frac{1}{x}/x^2$, you get $x^2f'(x) \rightarrow -1$ as $x \rightarrow \infty$. Hence the winning sandwich for your expression. (This may be a little similar to what is done when proving de l'Hopital but doesn't use the full force of it)


By the mean value theorem,

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

for some $c_x$ between $1/x$ and $1/(x-1).$ Thus as $x\to \infty, c_x\to 0,$ hence $e^{c_x}\to 1.$ Multiplying by $x^2$ then gives a limit of $1.$

  • $\begingroup$ This is L'Hospital without L'Hospital. $\endgroup$
    – yo'
    Mar 2, 2018 at 17:55
  • $\begingroup$ @yo' The MVT is not L'Hopital. $\endgroup$
    – zhw.
    Mar 2, 2018 at 17:59

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