# Finding the limit of an expression involving three types of functions

Rasmus

## Homework Statement

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

$$\lim_{x\to 0} \frac{ln(1 + sin^2x)}{e^{2x} - 1}$$

## 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.

$$\lim_{x\to 0} \frac{ln(1 + sin^2x)}{e^{2x} - 1}$$

First multiply by $$\frac{sin^2x}{sin^2x}$$

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

$$\frac{sin^2x}{e^{2x}}$$

is multiplied by $$\frac{x^2}{x^2}$$

which gives the standard limit $$\frac{sin x}{x}$$ squared, which also approaches 1 as x approaches zero.

$$\frac{x^2}{e^{2x} - 1}$$

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

$$\frac{2x}{e^{2x} - 1}$$

and

$$\frac{x}{2}$$

Giving the final expression

$$\lim_{x\to 0} \frac{ln(1 + sin^2x)}{sin^2x} \cdot \frac{sin^2x}{x^2} \cdot \frac{2x}{e^{2x} - 1} \cdot \frac{x}{2}$$

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
$$ln(1+\sin ^2x)^\frac{1}{e^{2x}-1}$$
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:

$\frac{1}{e^{2x} - 1} \cdot ln(1 + sin^2 x) = ln(1 + sin^2 x)^{\frac{1}{e^{2x} -1}}$

When x approaches 0

$ln(1 + sin^2 0)^{\frac{1}{e^{0} - 1}} = ln(1 + 0)^{\frac{1}{0}} = ln(1)^{∞} = ln(1) = 0$

Correct?