- #1
bob1182006
- 492
- 1
Homework Statement
[tex]\lim_{(x,y) \rightarrow (0,0)} \frac{x^2+sin^2 y}{2x^2+y^2}[/tex]
Homework Equations
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:
[tex]\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}[/tex]
and I work on the one on the left.
[tex]\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[/tex]
[tex]\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[/tex]
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?