# Did I do anything wrong in this limit evaluation?

1. Mar 8, 2012

### pylauzier

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

A problem in my book asks the reader to evaluate lim x→0 xln(2x) using L'Hôpital.

2. Relevant equations

None

3. The attempt at a solution

I think I got it right but I simply wanted to make sure I did no illegal manipulations (I'm self-studying this stuff).

I started by substituting u = 1/x to get an undeterminate form suitable for L'Hôpital. By doing so, lim x→0 becomes lim u→∞

lim x→0 xln(2x) = lim u→∞ ln(2/u) / u

Applying L'Hôpital:

= lim u→∞ (-u)/2u2

= lim u→∞ (-1)/2u

= 0

I know that's the right answer, but did I make any mistakes? Also, was I allowed to switch lim x→0 by lim u→∞? Thanks in advance!

2. Mar 8, 2012

### Dick

Sure, it's right. You didn't even really need a substitution. Write the original limit as ln(2x)/(1/x) which is equally indeterminant. Then l'Hopital gives (1/x)/(-1/x^2)=(-x). That's still 0.

3. Mar 8, 2012

### pylauzier

Oh right, that sure is simpler. Thanks again!