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Prove that $\lim_{t\rightarrow0^{+}}\frac{e^{-\frac{1}{t}}}{t^{k}}=0 $ for every positive integer $k$. I am trying to show that the function $f:\mathbb{R}\rightarrow\mathbb{R}$ defined by $f(t)=\begin{cases} \exp\left(-\frac{1}{t}\right), & \mbox{if }t>0\\ 0, & \mbox{if }t\leq0 \end{cases}$ is smooth and to prove this, I have to prove that the limit exists. I am having trouble showing that it exists. I know that it can be shown using L'Hospital's Rule, but i would prefer to show it using the $\epsilon - \delta $ definition.

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    $\begingroup$ Hint: Can you show that $x^{-k}e^x\to\infty$ when $x\to+\infty$? $\endgroup$ – Did Mar 19 '17 at 7:39
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Note that the exponential satisfies the inequality

$$\begin{align} e^x&\ge 1+x+\cdot +\frac{x^{k+1}}{(k+1)!}\\\\ &>\frac{x^{k+1}}{(k+1)!}\tag 1 \end{align}$$

Using $(1)$, it is easy to see that

$$\begin{align} \frac{e^{-1/t}}{t^k}&=\frac{1}{t^ke^{1/t}}\\\\ &\le \frac{1}{t^k\left(\frac{1}{(k+1)!t^{k+1}}\right)}\\\\ &=(k+1)!t \end{align}$$

Hence, given $\epsilon>0$, $\frac{e^{-1/t}}{t^k}<\epsilon$ whenever $0<t<\delta=\epsilon/(k+1)!$.

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Pick $\epsilon > 0$. By taking the derivative one sees that $e^{-1/t}t^{-k}$ is increasing on some interval $(0,a)$. Choose $m \in \mathbb{N}$ such that $2^{-m} < a$ and $$mk - 2^m < \log_2\epsilon.$$ Therefore if $0 < t < 2^{-m}$ we have $$\frac{e^{-1/t}}{t^k} < \frac{2^{-1/t}}{t^k} < \frac{2^{mk}}{2^{2^m}} = 2^{mk - 2^m} < \epsilon.$$

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