# Proving function Converges

1. Oct 9, 2013

### PsychonautQQ

1. The problem statement, all variables and given/known data
How do I prove the function ((2^x)-1)sin(y))/(xy) converges to ln(2) in the case of x=0 and y=0..?

2. Relevant equations

3. The attempt at a solution
Yeah I'm trying to figure out where to start honestly... I know i'm not suppose to post unless I have a real attempt at the solution.. I could BS something but anyone wanna point me in the right direction? Squeeze Theorem? Is there some key property I should be aware of?

2. Oct 9, 2013

### LCKurtz

Why do you think the answer is $\ln 2$? Try some special limits to get a feel for it. What if $y\rightarrow 0$ before $x$, for example.

3. Oct 9, 2013

### PsychonautQQ

the answer is ln(2), the question says that, we just have to PROVE it is.

4. Oct 10, 2013

### dirk_mec1

Yes, the answer is indeed ln(2) I used mclaurin series to prove it.

5. Oct 10, 2013

### pasmith

Your function is a function of x only times a function of y only:
$$\frac{(2^x - 1)\sin y}{xy} = \frac{2^x - 1}{x} \frac{\sin y}{y}$$

One can show that if $\lim_{x \to 0} f(x)$ and $\lim_{y \to 0} g(y)$ exist then
$$\lim_{(x,y) \to 0} f(x)g(y) = \left(\lim_{x \to 0} f(x) \right)\left( \lim_{y \to 0} g(y) \right)$$

Also remember that
$$\lim_{x \to 0} \frac{f(x) - f(0)}x = f'(0)$$
assuming the limit exists.