Limits: Calculating Along y=2x

[Cpt Qwark]

Homework Statement

Calculate \lim_{(x,y)\to(0,0)}\frac{x^4-4y^2}{x^2+2y^2} along the the line y=2x

Homework Equations

N/A

The Attempt at a Solution

Not too sure what they mean by calculating the limit along the line y=2x. The answer is \frac{-3}{5}.
But I have gotten so far: \lim_{(0,y)\to(0,0)}\frac{-y^2}{y^2}=-1 and \lim_{(x,0)\to(0,0)}\frac{x^2}{x^2}=1, but the limit doesn't exist l_1\neq{l_2}?

 

[haruspex]

Taking the limit along the line y=2x just means you can substitute y=2x then take the limit as x tends to 0.
Your two attempts took the limits along the lines x=0 (first attempt), y=0 (2nd attempt).

 

[Cpt Qwark]

Sorry, it was supposed to be \lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}, so you sub in y=2x and compute \lim_{x\to0}\frac{x^2-(2x)^2}{x^2+(2x)^2}?

 

[Mark44]

Sorry, it was supposed to be \lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}, so you sub in y=2x and compute \lim_{x\to0}\frac{x^2-(2x)^2}{x^2+(2x)^2}?

Yes.

Side note: Don't use BBCodes inside of LaTeX code. Your BBCode italics tags broke $l_1 \neq l_2$ in your first post.

 

[SammyS]

\lim_{x\to0}\frac{x^2-(2x)^2}{x^2+(2x)^2}?

Were you able to evaluate this limit ?

 

[Cpt Qwark]

Were you able to evaluate this limit ?

Yes.