When $t$ goes to infinity in function $-2 (te^{-t} + e^{-t} )$

Question

How to compute the following limit:

$$\lim_{t\to\infty}-2\left(te^{-t}+e^{-t}\right)$$

Answer 1

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: $$\begin{align} &\lim_{t\to\infty}-2\left(te^{-t}+e^{-t}\right)\\ &=-2\lim_{t\to\infty}\frac{\frac{\text{d}}{\text{d}t}t}{\frac{\text{d}}{\text{d}t}e^{t}}\\ &=-2\lim_{t\to\infty}\frac{1}{e^{t}}\\ &=0 \end{align}$$

Answer 2

$$\lim\limits_{t\to \infty}-2(\frac{t}{e^t}+e^{-t})$$

$$=-2\bigg(\lim\limits_{t\to \infty}e^{-t}+\lim\limits_{t\to \infty}e^{-t}t\bigg)$$

$$=-2\bigg(0+\lim\limits_{t\to \infty}e^{-t}t\bigg)$$

$$=-2\lim\limits_{t\to \infty}e^{-t}t$$

Since the polynomial $t$ grows asymptoticaly slower then $e^t$ as $t$ approaches $\infty$

$$\boxed{\color{red}{=-2\times 0=0}}$$

Answer 3

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 $t$.

Answer 4

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 $$=-2\lim_{x\to \infty}\left(\frac{\frac{d}{dt}(t+1)}{\frac{d}{dt}(e^t)}\right)$$

$$=-2\lim_{x\to \infty}\left(\frac{1}{e^t}\right)=0$$