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

Quote:

Originally Posted by

**Cotty** Hi,

I have this question:

lim (e^t-1+t)/t^2

t->0

so I have used L' Hopital's Rule because when you sub in 0 you get "0/0" and arrived at

lim e^t+1/2t

t->0

which i assume i would use L' Hops again if the original question was e^t-1-t/t^2 (as this derived would again give the answer "0/0") but i am not sure how to get the solution because I assume i cannot use L' Hops again because when 0 is subbed in you get "2/0". I did try this before and got the wrong answer, it says that the answer is infinity.

Thank you very much for your time!

After using L'Hospital's Rule once to get to $\displaystyle \displaystyle \lim_{t \to 0} \frac{e^t + 1}{2t} $, the function is no longer of the form $\displaystyle \displaystyle \frac{0}{0}$. The top is nonzero while the bottom goes to 0, which means the limit is $\displaystyle \displaystyle \infty $.

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

Ahh ok thank you, I just wasn't sure whether it would have to be manipulated more but I can see it clearly now that you have pointed it out, thanks.