Mampac
New member
- Joined
- Nov 20, 2019
- Messages
- 48
Hi, I have the following task:
When evaluating double limits I should try my best to get rid of discontinuity that appears once x and y are plugged in; I have to simplify this so that the denominator isn't 0.
OK, I used the [imath]\frac{sinx}{x}[/imath] useful limit to replace the denominator with [imath]2x^2+3y^2[/imath] (by multiplying and dividing by it).
Now I get [math]\lim_{(x, y) \to (0, 0)} \frac{xy\tan (x + y)}{2x^2+3y^2}[/math] which can be simplified further knowing that tangent is equal to sine divided by cosine, and applying the same property mentioned above, to get [math]\lim_{(x, y) \to (0,0)} \frac{x^2y +xy^2}{\cos(x+y)(2x^2+3y^2)}[/math] I'm stuck at this point. No obvious factoring can be done. I doubt it that switching to polar coordinates will do the trick. Any ideas what to do next?
Thank you in advance
When evaluating double limits I should try my best to get rid of discontinuity that appears once x and y are plugged in; I have to simplify this so that the denominator isn't 0.
OK, I used the [imath]\frac{sinx}{x}[/imath] useful limit to replace the denominator with [imath]2x^2+3y^2[/imath] (by multiplying and dividing by it).
Now I get [math]\lim_{(x, y) \to (0, 0)} \frac{xy\tan (x + y)}{2x^2+3y^2}[/math] which can be simplified further knowing that tangent is equal to sine divided by cosine, and applying the same property mentioned above, to get [math]\lim_{(x, y) \to (0,0)} \frac{x^2y +xy^2}{\cos(x+y)(2x^2+3y^2)}[/math] I'm stuck at this point. No obvious factoring can be done. I doubt it that switching to polar coordinates will do the trick. Any ideas what to do next?
Thank you in advance