Question

Limit question.

lim t->0 of (e^t+t-1)/t^2

Is this a question that cannot be evaluated with limit tricks, if so, how?

Answer 1

\[\lim_{t \rightarrow 0} \frac{e^t+t-1}{t^2}\]

Answer 2

There's l'Hospital's rule... does that count as a limit trick?

Answer 3

I would get:
\[\lim_{t \rightarrow 0} \frac{e^t+1}{2t}\]

When evaluated you get 2/0

Should I try do L'Hopital's rule again?

Answer 4

You can't do l'Hopital's again because it's not indeterminate form.

Answer 5

That is what I was thinking

Answer 6

I would check if it actually exists at this point.

Answer 7

I know the answer is infinity. I just don't know how/why

Answer 8

You sure about that?

Answer 9

For small negative values it tends towards \(-\infty\) doesn't it?

Answer 10

I hope so, if not that means my book is wrong, and also wolframalpha.

Answer 11

http://www.wolframalpha.com/input/?i=lim+t-%3E0+of+%28e^t%2Bt-1%29%2Ft^2

Answer 12

Yeah, when you have an asymptote that means the limit doesn't exist bro.

Answer 13

Cool. So how do I know there is an asymtope just by looking at the equation?

Answer 14

\[
\lim_{t\to 0^+}f(t) = \infty \neq -\infty =\lim_{t\to 0^-}f(t)
\]

Answer 15

When you suspect a limit doesn't exist, your only choice is really to plug in really close values.

Answer 16

For example.... plug in 0+(000.1) and 0-(000.1)

Answer 17

Ah, okay.

Thanks I guess.

^_^

Answer 18

Really, you underestimate how magical limits are...

Answer 19

There is more magic in limits then in all of Equestria.
If you could do a limit as trivially as you could do multiplication, most of calculus would be a joke.

Answer 20

Haha, I guess so. Thanks for your help.