I'm struggling with calculating the limit of the following function: $$\lim_{x\to 0^-}\frac{e^{\frac{1}{x}}}{x}$$ I tried using L'Hopital's rule, but I'm still getting $\frac{0}{0}$ and the denominator keeps getting greater exponent.
Limit of a function with e.
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limits
functions
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0write $$y=\frac{e^{1/x}}{x}$$ and take the logarithm – 2017-01-30
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0A plot suggests that the left limit is 0 and the right limit infinite. I guess l'Hospital doesn't work because those two are different. – 2017-01-30
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0@ Dr. Sonnhard Graubner Just asking. How can you take logarithm? Am I right to say that $y<0$ since $x\to 0^-?$ – 2017-01-30
1 Answers
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Change $t=-1/x$, so the limit becomes $$ \lim_{t\to\infty}\frac{e^{-t}}{-1/t}= \lim_{t\to\infty}-\frac{t}{e^t} $$