You are not logged in.
Pages: 1
here is a clear image of the limit
http://img35.imageshack.us/i/60400746.png/
I meant 1/3x,
however ,I think the answer is not infinity. Infinity is not one of the options in the possible multiple choices. I think l'hopital might have to be used. However, as I said, I am having trouble getting it into the indeterminate form
The possible answers are :
a)0
b)1
c)e^7
d)e^(2/3)
e)7
Hi,
I am having trouble with this limit, and I would appreciate anyone's imput.
limit (1+(1/3X))^(7+2X)
as x->0
I have tried to manipulate this into the 2 indeterminate forms that work with L'hopitals rule to no avail.
I am leaning toward the limit approaching 0 , since the limit is undefined at x=0, if we approach x from 0+, the limit goes to infinity^7 , which means 1/[(inifinity)^-7)] = 0
Thank you for your help
Thank you so much. You have been very helpful
Thank you for your quick reply.
As X approaches to 1 from above, the limit goes to + infinity and as X approaches to 1 from below, the limit goes to - infinity. Therefore it does not exist as X -> 1
What happens to the limit of F(x) as x approaches to 1:
[(x^2)-x-2]/[(x^2)-1]
L'hopital rule can not be applied in this case, since it is not 0/0 , and if we simply , we get (x-2)/(x-1), which doesn't work as well
I'm leaning towards that this limit does not exist. I'd be grateful if anyone can help me out
Pages: 1