# 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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### Jon

Maybe I'm a lion?

Updated on October 20, 2022

• 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?

• why did you take the derivative?
• This 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 $t\to 0$.
• So 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?
• Nah. 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.
• • • Exactly. Since the $j$ and $k$ components are continuous at $0$, you can just plug in $t=0$ to evaluate those particular limits, leaving $\frac{1}{2}i+j$.
• @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.