How to compute the following limit:
$$\lim_{t\to\infty}-2\left(te^{-t}+e^{-t}\right)$$
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There are many ways of computing this limit. One among the easiest ones is probably the following, using l'Hospital's rule:
$$\begin{align} &\lim_{t\to\infty}-2\left(te^{-t}+e^{-t}\right)\\ &=-2\lim_{t\to\infty}\left(te^{-t}+e^{-t}\right)\\ &=-2\lim_{t\to\infty}\frac{t+1}{e^{t}}\\ &=-2\lim_{t\to\infty}\frac{t}{e^{t}}\\ &"=" \frac{\infty}{\infty} \end{align}$$ We can use l'Hospital's rule here: $$\begin{align} &\lim_{t\to\infty}-2\left(te^{-t}+e^{-t}\right)\\ &=-2\lim_{t\to\infty}\frac{\frac{\text{d}}{\text{d}t}t}{\frac{\text{d}}{\text{d}t}e^{t}}\\ &=-2\lim_{t\to\infty}\frac{1}{e^{t}}\\ &=0 \end{align}$$
MoebiusCorzer
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You can use this to find the result:
When $t\to\infty$, we have $te^{-t}\to 0$, because the exponential function wins over the factor $t$.
Mankind
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You cleared my doubt thanks – D_1027 Dec 11 '15 at 13:33
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@DushaniWellappili It is better to give a proper argument, rather than "because the exponential function wins over the factor $t$" if you want to properly compute the integral. Of course, if you just want some intuitive reasoning, this is correct. – MoebiusCorzer Dec 11 '15 at 13:36
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Notice, $$\lim_{x\to \infty}-2(te^{-t}+e^{-t})$$ $$=-2\lim_{x\to \infty}e^{-t}(t+1)$$ $$=-2\lim_{x\to \infty}\left(\frac{t+1}{e^t}\right)$$ Applying L'Hopspital's rule for $\frac{\infty}{\infty}$ form $$=-2\lim_{x\to \infty}\left(\frac{\frac{d}{dt}(t+1)}{\frac{d}{dt}(e^t)}\right)$$
$$=-2\lim_{x\to \infty}\left(\frac{1}{e^t}\right)=0$$
Harish Chandra Rajpoot
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$$\lim\limits_{t\to \infty}-2(\frac{t}{e^t}+e^{-t})$$
$$=-2\bigg(\lim\limits_{t\to \infty}e^{-t}+\lim\limits_{t\to \infty}e^{-t}t\bigg)$$
$$=-2\bigg(0+\lim\limits_{t\to \infty}e^{-t}t\bigg)$$
$$=-2\lim\limits_{t\to \infty}e^{-t}t$$
Since the polynomial $t$ grows asymptoticaly slower then $e^t$ as $t$ approaches $\infty$
$$\boxed{\color{red}{=-2\times 0=0}}$$
3SAT
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