Calculate the limit of a vector?
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Solution 1
Note that $$ \lim_{t\to0}\frac{\sin t}{2t} = \frac{1}{2}. $$ (Think L'Hopital's!)
Solution 2
You should use L'Hospital Rule for $i$ component only, getting $\frac12$. For $j$ component it goes to $1 $, and $k$ component limit is clearly zero.
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Comments

Jon about 1 year
Find $$~\lim_{t→0} f (t)~$$ if it exists
$$f(t) = \dfrac{\sin t}{2t} ~\hat i + \, e^{2 t}~\hat j + \dfrac{t^2}{e^{t}}~\hat k ~.$$
When I plug in $~t\to 0~$ for $~f(t)~$, I get $0+j+0$.
But the answer is $\left( \frac{1}{2} ~\hat i + ~\hat j\right)$.
Could someone tell me where I have made grievous mistake? 
Jon about 8 yearswhy did you take the derivative?

Jesse about 8 yearsThis is just the limit as $t$ approaches $0$ of the $i$ component of your vector $f(t)$. It can be calculated using derivatives via L'Hopital's Rule or by using the known limit $\frac{\sin t}{t}\to 1$ as $t\to 0$.

Jon about 8 yearsSo you did L'Hopital's rule because the function was approaching 0. So therefore I am to assume I take L'Hopital's rule for the rest yes?

Jesse about 8 yearsNah. Remember that L'Hopital's applies when taking limits of indeterminate forms, such as $\frac{0}{0}$ (which is what this limit was), $\frac{\infty}{\infty}$ and so on.

Jon about 8 yearsI see for the rest you just plug in lim t=> 0. and get 1/2i+j.

Jon about 8 yearsBut moreover this is basically 0/0i + 1 + 0/1. Since it is 0/1 l'hospitals rule is not required.

Jesse about 8 yearsExactly. Since the $j$ and $k$ components are continuous at $0$, you can just plug in $t=0$ to evaluate those particular limits, leaving $\frac{1}{2}i+j$.

Mark Viola about 8 years@jesse To be precise, L'Hospital's Rule applies to cases for which just the denominator goes to $\infty$. This fact is a consequence of Cesaro Stolz and is often not taught at intro levels.