1. ## Vector Function Help

I need some help understanding how:

Find ${\lim_{t \to 0} r(t)}$, where $r(t) = (1+t^3)$ i + $(te^{-t})$ j + $(\frac{sin(t)}{t})$ k

= ${\lim_{t \to 0} r(t)} = [{\lim_{t \to 0} (1+t^3)}]$ i + $[{\lim_{t \to 0} (te^{-t})}]$ j + $[{\lim_{t \to 0} (\frac{sin(t)}{t})}]$ k

= i + k (form the book)

I would think that one would have to replace the t by 0, getting:

= ${\lim_{t \to 0} r(t)} = [{\lim_{t \to 0} (1+0^3)}]$ i + $[{\lim_{t \to 0} ((0)e^{0})}]$ j + $[{\lim_{t \to 0} (\frac{sin(0)}{0})}]$ k

= ${\lim_{t \to 0} r(t)} = [{\lim_{t \to 0} (1)}]$ i + $[{\lim_{t \to 0} (0)}]$ j + $[{\lim_{t \to 0} (D.N.E)}]$ k

= i

So I guess what I'm asking is why is it equal to i + k and not i?

2. $[{\lim_{x \to 0} (\frac{sin(x)}{x})}] = 1$.

It's just one of those things, a "special limit" to basically just memorize. At least that's how I look at it.

If you want to see a proof of this particular limit, this is a good place to start:

http://www.csun.edu/~ac53971/courses.../xtra_sine.pdf.

Also, if a limit does not exist, as you posited for your k vector, that doesn't mean it's zero - zero is a number that exists just like 1, 2, 3, etc. If you find that a limit does not exist but you need to evaluate it, you need to figure out a way TO evaluate it.

3. Thanks, but what if you have other values like:

1) ${\lim_{t \to 0} \frac{e^{t}-1}{t}}$

2) ${\lim_{t \to 0} \frac{t^2}{sin^{2}t}}$

3) ${\lim_{t \to 0} \frac{\sqrt{1+t}-1}{t}}$

will these be equal to some specific number, or will they all be undefined? And or those specific properties only applicable on trigonometric functions?

4. Originally Posted by lllll
Thanks, but what if you have other values like:

1) ${\lim_{t \to 0} \frac{e^{t}-1}{t}}$

2) ${\lim_{t \to 0} \frac{t^2}{sin^{2}t}}$

3) ${\lim_{t \to 0} \frac{\sqrt{1+t}-1}{t}}$

will these be equal to some specific number, or will they all be undefined? And or those specific properties only applicable on trigonometric functions?
There are a number of different approaches that can be taken.

1) From l'Hopital's rule:

$\lim_{t \rightarrow 0} \frac{e^t - 1}{t} = \lim_{t \rightarrow 0} \frac{e^t}{1} = 1$.

You could also substitute the power series for $e^t$:

$\lim_{t \rightarrow 0} \frac{e^t - 1}{t} = \lim_{t \rightarrow 0} \frac{(1 + t + \frac{t^2}{2!} + ...) - 1}{t} = \lim_{t \rightarrow 0} (1 + \frac{t}{2!} + ....) = 1$.

2) From l'Hopital's Rule:

$\lim_{t \rightarrow 0} \frac{t^2}{\sin^2 t} = \lim_{t \rightarrow 0} \frac{2t}{2 \sin t \cos t} = \lim_{t \rightarrow 0} \frac{2t}{\sin (2t)} = \lim_{t \rightarrow 0} \frac{2}{2 \cos (2t)} = 1$.

Alternatively, using known limits:

$\lim_{t \rightarrow 0} \frac{t^2}{\sin^2 t} = \lim_{t \rightarrow 0} \left( \frac{t}{\sin t} \right) \lim_{t \rightarrow 0} \left( \frac{t}{\sin t} \right) = 1 \times 1 = 1$.

3) $\lim_{t \rightarrow 0} \frac{\sqrt{1+t} - 1}{t} = \lim_{t \rightarrow 0} \frac{(\sqrt{1+t} - 1)}{t} \times \frac{(\sqrt{1+t} + 1)}{(\sqrt{1+t} + 1)}$

$= \lim_{t \rightarrow 0} \frac{(1 + t) - 1}{t (\sqrt{1+t} + 1)} = \lim_{t \rightarrow 0} \frac{1}{\sqrt{1+t} + 1} = \frac{1}{1 + 1} = \frac{1}{2}$.