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How can I evaluate \[ \lim_{t\to 0} \frac{e^{-1/t}}{t}\quad ? \] I tried to use L'Hôpital's rule but it didn't help me. Any hints are welcome. Thanks.

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3 Answers 3

up vote 6 down vote accepted

Note that as $t \to 0$, $\exp(-1/t)$ tends 'faster' to $0$ than $1/t$ tends to $\infty$. To make this precise, let us proceed as follows.

First note that $\exp(x) \geq 1 + x + \dfrac{x^2}2$ for $x \geq 0$ and $\exp(x) \leq 1 + x + \dfrac{x^2}2$ for $x \leq 0$. For $t \geq 0$, $$\dfrac{\exp(-1/t)}{t} = \dfrac1{t \exp(1/t)} \leq \dfrac1{t (1 + 1/t + 1/(2t^2))} = \dfrac1{t + 1 + 1/(2t)} = \dfrac{2t}{2t^2 + 2t + 1}$$ Hence, if we let $t \rightarrow 0^+$, we get that $$0 \leq \lim_{t \rightarrow 0^+} \dfrac{\exp(-1/t)}{t} \leq \lim_{t \rightarrow 0^+} \dfrac{2t}{2t^2 + 2t + 1} = 0$$ Argue similarly, for $t \leq 0$.

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@Jr. $\lim_{t \rightarrow a} \dfrac{f(t)}{g(t)} = \dfrac{f(a)}{g(a)}$ whenever $f$ and $g$ are nice functions and when $f(a)$, $g(a)$ makes sense. –  user17762 Jun 9 '12 at 18:12
    
surely, thanks :) –  Jr. Jun 9 '12 at 18:14

You are given

$$\lim_{t \to 0} \frac{1}{t}\exp\left({-\frac 1 t}\right)$$

Since $1/t$ behaves oppositely for $0^+$ or $0^-$, we consider both situations. Then we let $x =\dfrac 1 t $ and get

$$\lim_{t \to 0^+} \frac{1}{t}\exp\left({-\frac 1 t}\right)=\lim_{x \to+\infty}xe^{-x}=\lim_{x \to+\infty}\frac x {e^{x}}$$

$$\lim_{t \to 0^-} \frac{1}{t}\exp\left({-\frac 1 t}\right)=\lim_{x \to -\infty}xe^{-x}$$

I guess calculation is now straightforward.

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Hint: Set $x=1/t$. (Also, you need to consider $t \to 0^+$ and $t \to 0^-$ separately.)

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