# Unfamiliar algebra within L'Hopital

1. Mar 12, 2012

### LearninDaMath

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

$\stackrel{lim}{x\rightarrow}0^{+}$ ($e^{2x}-1)^{x}$

At the step of dividing by e^x, how does that algebra work?

2. Mar 12, 2012

### stripes

e^(2x) was factored out from the denominator. In the step right before the blue box, we can factor out e^(2x) from (e^(2x) - 1) so we have [e^(2x)](1 - (1/e^(2x))) = [e^(2x)](1 - e^(-2x)):

$\frac{e^{2x}}{e^{2x} - 1} = \frac{e^{2x}}{e^{2x}(1 - \frac{1}{e^{2x}})} = \frac{1}{1 - e^{-2x}}$.

3. Mar 12, 2012

### LearninDaMath

Thanks stripes, appreciate the illustration with the latex reference as well.

I suppose that the reason for factoring out and cancelling away e^2x from this rational function is to avoid having to do an unnecessary product rule (and or chain rule) when applying L'hopitals for the second time. Would this be right?

For future problems similar to this one, what should I be thinking when doing these problems so as to notice when something can be factored out in order to make the next steps easier? Should I just look for similar terms in the numerator and denominator, and if there are, just go ahead and cancel them away before proceeding to any next steps?

Last edited: Mar 12, 2012
4. Mar 12, 2012

### stripes

Yes, doing this eliminates the need for the chain rule in the numerator, so by cancelling the e^(2x), we are simplifying the numerator, and completely eliminating the denominator (i.e., turning the denominator to 1), and thereby making the entire expressing non-indeterminate form.

The best thing to do is to try to turn the expression into a rational one (a ratio of two expressions), and then trying to simplify the numerator if necessary. Sometimes you will be able to take the derivative of the top and bottom easily, but when you cannot, then try to manipulate the expression a little bit in your favor. Look for similar terms, anything that can lead to an expression that is easier to work with.

Unfortunately, depending on the problem, this might be easier said than done. But perseverance is key!

5. Mar 12, 2012

### LearninDaMath

Thanks, appreciate your help and advice. I will continue to persevere.