# Homework Help: Multivariable Limit, is my reasoning correct?

1. Sep 26, 2007

### bob1182006

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

$$\lim_{(x,y) \rightarrow (0,0)} \frac{x^2+sin^2 y}{2x^2+y^2}$$

2. Relevant equations

3. The attempt at a solution

Since I can't evaluate the limit directly and I can't see a way to get a 0 on the top in order to get 2 different limits I split up the limit into these 2:
$$\lim_{(x,y) \rightarrow (0,0)} \{\frac{x^2}{x^2+y^2}+\frac{sin^2}{x^2+y^2}\}=\lim_{(x,y) \rightarrow (0,0)} \frac{x^2}{2x^2+y^2}+\lim_{(x,y) \rightarrow (0,0)} \frac{sin^2 y}{2x^2+y^2}$$

and I work on the one on the left.

$$\lim_{\substack{x=0\\y\rightarrow 0}} \frac{x^2}{2x^2+y^2}=\lim_{y \rightarrow 0} \frac{0}{y^2}=\lim_{y \rightarrow 0} 0 = 0$$

$$\lim_{\substack{y=x\\x\rightarrow 0}} \frac{x^2}{2x^2+y^2}=\lim_{x \rightarrow 0} \frac{x^2}{3x^2}=\lim_{x \rightarrow 0} \frac{1}{3}=\frac{1}{3}\neq0$$

and since the left limit doesn't exist it doesn't matter if the right limit exists since the sum of a nonexisting limit and something else won't exist either, right?

2. Sep 26, 2007

### Avodyne

Not necessarily; the right limit might not exist, but the nonexistence might cancel between the two terms. For example, if the problem was (x^2+y^2)/(x^2+y^2), the limit is obviously 1, but the limits of x^2/(x^2+y^2) and y^2/(x^2+y^2) do not individually exist.

But in your case, when y is small, what happens to sin(y)? Just do your two examples with the full numerator.

Last edited: Sep 26, 2007
3. Sep 26, 2007

### bob1182006

oo because values near 0, sin(y)=y so it would be about the same method of evaluating!

thanks alot!