Find Double Limit of a Fraction Involving Trigonometric Functions

[Mampac]

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

 

[blamocur]

If [imath](x,y)[/imath] is close to [imath](0,0)[/imath] then [imath]\cos(x+y)[/imath] is close to 1.

 

[Mampac]

If [imath](x,y)[/imath] is close to [imath](0,0)[/imath] then [imath]\cos(x+y)[/imath] is close to 1.

I absolutely agree with you, I noticed it myself, but I cannot see how [math]lim_{(x,y) \to (0,0)} \frac{x^2y + xy^2}{2x^2+3y^2}[/math] can be simplified further!

 

[blamocur]

Not sure it needs to be simplified any further. I don't know what material from your class you can use (i.e., theorems and lemmas already proven), but I believe this can be solved using only the [imath](\epsilon,\delta)[/imath] definition of limit (https://demonstrations.wolfram.com/...eople.math.umass.edu/~havens/LimContBivar.pdf).

 

[BigBeachBanana]

I absolutely agree with you, I noticed it myself, but I cannot see how [math]lim_{(x,y) \to (0,0)} \frac{x^2y + xy^2}{2x^2+3y^2}[/math] can be simplified

I absolutely agree with you, I noticed it myself, but I cannot see how [math]lim_{(x,y) \to (0,0)} \frac{x^2y + xy^2}{2x^2+3y^2}[/math] can be simplified further!

One approach is to use Squeeze Theorem. Make sure that your upper and lower bound functions remain above and below the function respectively in a neighborhood around the point you are interested in.

Hi, I have the following task:
View attachment 30151
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

I'm not familiar with the tg(x+y) notation. Is this tangent of (x + y)?

 

[BigBeachBanana]

I absolutely agree with you, I noticed it myself, but I cannot see how [math]lim_{(x,y) \to (0,0)} \frac{x^2y + xy^2}{2x^2+3y^2}[/math] can be simplified further!

One approach is to use Squeeze Theorem. Make sure that your upper and lower bound functions remain above and below the function respectively in a neighborhood around the point you are interested in.

Hi, I have the following task:
View attachment 30151
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

I'm not familiar with the tg(x+y) notation in the original question. Is this tangent(x+y)?

 

[blamocur]

Like the idea of using Squeeze Theorem (https://en.wikipedia.org/wiki/Squeeze_theorem)

 

[Steven G]

Continuing from where you got to, let x=0 (or y=0) and the result will be obvious. Now use ϵ-δ definition of limit to confirm this limit