Find the limit of the vector function

Question

$lim_{t\to\infty} \Big(te^{-t},\frac{t^3+t}{2t^3-1},tsin(\frac{1}{t})\Big)$

a) $lim_{t\to\infty} te^{-t} = \infty \times 0$

$lim_{t\to\infty} 1e^{-t}+-e^tt = 0+(0\times\infty)$=undefined, and repeating l'hospitals rule will render the same result over and over.

b) $lim_{t\to\infty}\frac{t^3+t}{2t^3-1} = \frac{\infty}{\infty}$

$lim_{t\to\infty}6=6$ (simpflifying using l'hospitals rule)

c) $lim_{t\to\infty} tsin\Big(\frac{1}{t}\Big) = \infty \times 0$

$lim_{t\to\infty} 1sin\Big(\frac{1}{t}\Big)+cos\Big(\frac{1}{t}\Big)\frac{-t}{t^2}= \lim_{t\to\infty} sin\Big(\frac{1}{\infty}\Big)+cos\Big(\frac{1}{\infty}\Big)\frac{-1}{\infty} = 0+(1\times0)=0$

$lim_{t\to\infty}\Big(te^{-t},\frac{t^3+t}{st^3-1},tsin(\frac{1}{t})\Big) = \Big(idk,6,0\Big)$

Are b and c correct? I am not sure what to do about a and this is an even numbered exercise in my book so idk what the answer is.

Answer 1

(a) $$\lim_{t\to \infty}te^{-t}=\lim_{t\to \infty}\frac{t}{e^t}=\lim_{t\to \infty}\frac{1}{e^t}=0$$ (b) $$\lim_{t\to \infty}\frac{t^3+t}{2t^3-1}=\lim_{t\to \infty}\frac{1+1/t^2}{2-1/t^3}=\frac{1}{2}$$ (c) $$\lim_{t\to \infty}t\sin \frac{1}{t}=\lim_{t\to \infty}\frac{\sin\frac{1}{t}}{\frac{1}{t}}=\lim_{\frac{1}{t}\to 0}\frac{\sin\frac{1}{t}}{\frac{1}{t}}=1$$

(a) Using L'Hopital's rule in the second step.

(b) No need to use L'Hopital's rule, but if you want, it will be $$\lim_{t\to \infty}\frac{t^3+t}{2t^3-1}=\lim_{t\to \infty}\frac{3t^2+1}{6t^2}=\lim_{t\to\infty}\frac{6t}{12t}=\lim_{t\to\infty}\frac{6}{12}=\frac{1}{2}$$

(c) The last step is well known. But if you want L'Hopital's rule, it is $$\lim_{x\to 0}\frac{\sin x}{x}=\lim_{x\to 0}\frac{\cos x}{1}=1$$ where $x=1/t$.

Answer 2

You seem to be mistaken in your use of L'Hopital's Rule. To properly evaluate a limit of the form $0\cdot\infty$, you must convert it to an equivalent one of the form $0/0$ or $\infty/\infty$.

For the first limit, using $te^{-t}=t/e^t$ will work just fine.

For the second limit, we have a limit of proper form, but the derivatives appear to have been derived incorrectly. Using L'Hopital's rule once more yields

$$\lim_{t\to\infty} \frac{t^3+t}{2t^3-1}=\lim_{t\to\infty} \frac{3t^2+1}{6t^2}=\lim_{t\to\infty} \frac{6t}{12t}=\frac{1}{2}$$

For the final limit, we can create a proper form by using $t\sin(1/t)=(\sin(1/t))/(1/t)$. Thus,

$$\lim_{t\to\infty}t\sin(1/t)=\lim_{1/t\to0}\frac{\sin(1/t)}{1/t}=1$$