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What is the limit $$\lim\limits_{x\rightarrow\infty} e^{\,x-1} ?$$

Thanks.

Thomas
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ypp
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  • Try plugging in larger and larger values on x on your calculator. What is $e^{100-1}$? $e^{1000-1}$? $e^{10000-1}$, etc. Also, try typing \limits before the _ to get the $x\rightarrow\infty$ under the $\lim$ like this, $\lim\limits_{x\rightarrow\infty}$ – Jürgen Sukumaran Jun 21 '16 at 17:27
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    Are you asking for the value of the limit $\lim_{x\to \infty}e^{x-1}$? Or are you asking for the value of the limit $\lim_{x\to \infty}e^{x^{-1}}$? – Mark Viola Jun 21 '16 at 17:27
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    The edits have already MathJax'ed your question, but maybe not correctly? (@Dr.MV's question points at another interpretation of what you wrote. To avoid things like that happening (and just because it makes math look much nicer than the alternatives), you should learn LaTeX/MathJax, you could start with MathJax help – Henrik supports the community Jun 21 '16 at 17:40
  • In the future, you may want to add context to your question. What do you suspect the limit is? Why aren't you confident in what you think? Are you proving this, or do you just need to know the limit? etc etc – pjs36 Jun 21 '16 at 18:41

3 Answers3

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As $x$ tends to infinity, $x-1$ goes toward $+\infty$, therefore: $$\lim_{x\to+\infty}e^{x-1}=\lim_{x\to+\infty}e^x.$$

C. Falcon
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$$\lim_{x\rightarrow \infty}e^{x-1}=\lim_{x\rightarrow \infty}\frac{e^{x}}{e}=\frac{1}{e}\lim_{x\rightarrow \infty}e^{x}=+\infty$$

operatorerror
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In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that

$$e^x\ge 1+x \tag 2$$

for all $x$. Therefore, we have

$$e^{x-1}\ge x$$

and hence

$$\lim_{x\to \infty}e^{x-1}\ge \lim_{x\to \infty}x=\infty$$

Community
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Mark Viola
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