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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?

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

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  • $\begingroup$ There must be some error, since the limit should be 1. $\endgroup$ – AlexanderJ93 Mar 2 '18 at 13:23
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    $\begingroup$ @AlexanderJ93 Thank you! I fixed. See now. $\endgroup$ – Michael Rozenberg Mar 2 '18 at 13:25
  • $\begingroup$ Isn't the stacked fraction on the left in the last bracket essentially "0/0"? $\endgroup$ – Max Freiburghaus Mar 2 '18 at 16:28
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    $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Mar 2 '18 at 16:32
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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 $$\frac{y}{1-y}=y+y^2+O\left(y^3\right)$$ $$e^y=1+y+\frac{y^2}{2}+O\left(y^3\right)$$ $$e^{\frac{y}{1-y}}=1+y+\frac{3 y^2}{2}+O\left(y^3\right)$$ $$e^{\frac{y}{1-y}}-e^y=y^2+O\left(y^3\right)$$

Edit

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 $\approx 1.020304224$

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By first order expansion for exponential

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

then

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

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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)

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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 $1.$

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  • $\begingroup$ This is L'Hospital without L'Hospital. $\endgroup$ – yo' Mar 2 '18 at 17:55
  • $\begingroup$ @yo' The MVT is not L'Hopital. $\endgroup$ – zhw. Mar 2 '18 at 17:59

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