Finding the limit of an expression involving three types of functions

[Rasmus]

Homework Statement

Find the limit (without using L'Hospital's rule)

[tex]\lim_{x\to 0} \frac{ln(1 + sin^2x)}{e^{2x} - 1}[/tex]

Homework Equations

The Attempt at a Solution

I tried various substitutions in order to rewrite the expression to a standard limit. Such as t = e^x, t = ln(x), t = sin(x) as well as using rewriting the trigonometric expression a few times. Since I suck I didn't get towards anything I recognized.

I'd be grateful for any help sucking slightly less :)

 

[Saitama]

Can you convert it to the form 1^infinity?

 

[Rasmus]

Can you convert it to the form 1^infinity?

Not a clue :)

I did however get help with a solution elsewhere.

[tex]\lim_{x\to 0} \frac{ln(1 + sin^2x)}{e^{2x} - 1}[/tex]

First multiply by [tex]\frac{sin^2x}{sin^2x}[/tex]

Then [tex]\frac{ln(1 + sin^2x)}{sin^2x}[/tex] is on standard form and will approach one as x approaches zero.

[tex]\frac{sin^2x}{e^{2x}}[/tex]

is multiplied by [tex]\frac{x^2}{x^2}[/tex]

which gives the standard limit [tex]\frac{sin x}{x}[/tex] squared, which also approaches 1 as x approaches zero.

[tex]\frac{x^2}{e^{2x} - 1}[/tex]

Is multiplied by 2/2 which gives the remaining factors as standard limits.

[tex]\frac{2x}{e^{2x} - 1}[/tex]

and

[tex]\frac{x}{2}[/tex]

Giving the final expression

[tex]\lim_{x\to 0} \frac{ln(1 + sin^2x)}{sin^2x} \cdot \frac{sin^2x}{x^2} \cdot \frac{2x}{e^{2x} - 1} \cdot \frac{x}{2}[/tex]

Since the first three factors approaches 1, and the fourth approaches 0, as x approaches 0 the limit will be 0.

 

[Saitama]

That's a good method.
What i meant was writing the given expression as
[tex]ln(1+\sin ^2x)^\frac{1}{e^{2x}-1}[/tex]
Now, its of the form 1^infinity in the logartihm.

 

[Rasmus]

Edit: Nevermind I misunderstood

I don't really understand how that expression evaluates though.

 

[Rasmus]

Here's my thinking:

[itex]\frac{1}{e^{2x} - 1} \cdot ln(1 + sin^2 x) = ln(1 + sin^2 x)^{\frac{1}{e^{2x} -1}}[/itex]

When x approaches 0

[itex]ln(1 + sin^2 0)^{\frac{1}{e^{0} - 1}} = ln(1 + 0)^{\frac{1}{0}} = ln(1)^{∞} = ln(1) = 0[/itex]

Correct?