# limit problem

• Apr 21st 2011, 06:15 PM
frankinaround
limit problem
this question is the limit as t tends to infinity of T(ln(5+1/t) - ln(5))

I think im having an algebraic problem. basicly the problem can be arranged :

infinity times 0 =

(ln(5+1/t) - ln(5)) / 1/T

that makes it 0/0 so I can use L'hopitals.

so I take the derivatives and get this :

((1/(5+1/T))(-1/t^2) - 1/5) / (-1/t^2)

But im not sure what to do from here. Ive been trying different thigns but keep getting different answers. the right answers supposed to be 1/5. can someone break it down for me?
• Apr 21st 2011, 06:26 PM
Prove It
To start with, ln(5) is a constant. The derivative of a constant is 0. Fix that error and you should start to see things cancelling.
• Apr 21st 2011, 06:35 PM
chisigma
Quote:

Originally Posted by frankinaround
this question is the limit as t tends to infinity of T(ln(5+1/t) - ln(5))

I think im having an algebraic problem. basicly the problem can be arranged :

infinity times 0 =

(ln(5+1/t) - ln(5)) / 1/T

that makes it 0/0 so I can use L'hopitals.

so I take the derivatives and get this :

((1/(5+1/T))(-1/t^2) - 1/5) / (-1/t^2)

But im not sure what to do from here. Ive been trying different thigns but keep getting different answers. the right answers supposed to be 1/5. can someone break it down for me?

Take into account that for t 'large enough' is...

http://quicklatex.com/cache3/ql_4143...4e71ffd_l3.png (1)

... then multiplying (1) by t...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Apr 23rd 2011, 11:24 AM
frankinaround
Ohhhh. I always thought ln(x) was 1/x, but now I know its chain rule. So ln(5) = 1/5 times d/dx(5) so 1/5 times 0 = 0. right? really ? whoa.
• Apr 23rd 2011, 11:38 AM
topsquark
Quote:

Originally Posted by frankinaround
Ohhhh. I always thought ln(x) was 1/x, but now I know its chain rule. So ln(5) = 1/5 times d/dx(5) so 1/5 times 0 = 0. right? really ? whoa.

Well, technically yes. But (d/dx)(constant) = 0 automatically. There is no need for the chain rule.

-Dan