Find Double Limit of a Fraction Involving Trigonometric Functions

Mampac

New member
Joined
Nov 20, 2019
Messages
49
Hi, I have the following task:
1639310086564.png
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 sinxx\frac{sinx}{x} useful limit to replace the denominator with 2x2+3y22x^2+3y^2 (by multiplying and dividing by it).

Now I get lim(x,y)(0,0)xytan(x+y)2x2+3y2\lim_{(x, y) \to (0, 0)} \frac{xy\tan (x + y)}{2x^2+3y^2} which can be simplified further knowing that tangent is equal to sine divided by cosine, and applying the same property mentioned above, to get lim(x,y)(0,0)x2y+xy2cos(x+y)(2x2+3y2)\lim_{(x, y) \to (0,0)} \frac{x^2y +xy^2}{\cos(x+y)(2x^2+3y^2)} 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
 
If (x,y)(x,y) is close to (0,0)(0,0) then cos(x+y)\cos(x+y) is close to 1.
 
If (x,y)(x,y) is close to (0,0)(0,0) then cos(x+y)\cos(x+y) is close to 1.
I absolutely agree with you, I noticed it myself, but I cannot see how lim(x,y)(0,0)x2y+xy22x2+3y2lim_{(x,y) \to (0,0)} \frac{x^2y + xy^2}{2x^2+3y^2} can be simplified further!
 
I absolutely agree with you, I noticed it myself, but I cannot see how lim(x,y)(0,0)x2y+xy22x2+3y2lim_{(x,y) \to (0,0)} \frac{x^2y + xy^2}{2x^2+3y^2} can be simplified
I absolutely agree with you, I noticed it myself, but I cannot see how lim(x,y)(0,0)x2y+xy22x2+3y2lim_{(x,y) \to (0,0)} \frac{x^2y + xy^2}{2x^2+3y^2} 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 sinxx\frac{sinx}{x} useful limit to replace the denominator with 2x2+3y22x^2+3y^2 (by multiplying and dividing by it).

Now I get lim(x,y)(0,0)xytan(x+y)2x2+3y2\lim_{(x, y) \to (0, 0)} \frac{xy\tan (x + y)}{2x^2+3y^2} which can be simplified further knowing that tangent is equal to sine divided by cosine, and applying the same property mentioned above, to get lim(x,y)(0,0)x2y+xy2cos(x+y)(2x2+3y2)\lim_{(x, y) \to (0,0)} \frac{x^2y +xy^2}{\cos(x+y)(2x^2+3y^2)} 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)?
 
I absolutely agree with you, I noticed it myself, but I cannot see how lim(x,y)(0,0)x2y+xy22x2+3y2lim_{(x,y) \to (0,0)} \frac{x^2y + xy^2}{2x^2+3y^2} 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 sinxx\frac{sinx}{x} useful limit to replace the denominator with 2x2+3y22x^2+3y^2 (by multiplying and dividing by it).

Now I get lim(x,y)(0,0)xytan(x+y)2x2+3y2\lim_{(x, y) \to (0, 0)} \frac{xy\tan (x + y)}{2x^2+3y^2} which can be simplified further knowing that tangent is equal to sine divided by cosine, and applying the same property mentioned above, to get lim(x,y)(0,0)x2y+xy2cos(x+y)(2x2+3y2)\lim_{(x, y) \to (0,0)} \frac{x^2y +xy^2}{\cos(x+y)(2x^2+3y^2)} 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)?
 
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
 
Top