# Limit, and lopitals rule

1. May 16, 2009

### gayani

1. The problem statement, all variables and given/known data

Find A so that lim (x+A/x-2A)^x =5
x-> 00

2. Relevant equations

3. The attempt at a solution
gotta use lopihtals rule.

2. May 16, 2009

### tiny-tim

Welcome to PF!

Hi gayani! Welcome to PF!

(have an infinity: ∞ and try using the X2 tag just above the Reply box )
hmm … I can't see how to use l'Hôpital's rule

but if you write the LHS as (1 + 3A/(x - 2A))x, and fiddle about with it, you should get something you recognise.

3. May 16, 2009

### Staff: Mentor

Gayani,
Just in case you don't follow what Tiny-tim did, here it is.

You wrote the rational expression as x + A/x - 2A. What I'm pretty sure you meant (and the way Tiny-tim interpreted it) was (x + A)/(x - 2A). What you wrote would be correctly interpreted as the sum and difference of three expressions-- x, A/x, and 2/A, not the quotient of x + A and x - 2A.

What Tiny-tim did was divide x + A by x - 2A, so that (x + A)/(x - 2A) = 1 + 3A/(x - 2A).

4. May 16, 2009

### Bohrok

Take the natural log of both sides and then take the limit of the natual log of the left side

$$\lim_{x\rightarrow\infty} ln\left(1 + \frac{3A} {x-2A}\right)^x= ln (5)$$

You'll have to do some rearranging in order to use l'Hôpital's rule and "solve" for A.

Last edited: May 16, 2009