- #1

pylauzier

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## Homework Statement

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

## Homework Equations

None

## 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)/2u

^{2}

= 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!