# Math Help - Finding & Proving the limit of lim (x,y)->(0,0) sin(x-y)/cos(x+y)

1. ## Finding & Proving the limit of lim (x,y)->(0,0) sin(x-y)/cos(x+y)

It says in my textbook that $sin(x-y)/cos(x+y) = sin(0)/cos(0) = 0/1 = 0$, so 0 is the limit. For most of the other problems in the section, I have to show that when the limit exists using the definition of limits of two-variable functions (that sqrt(x^2 + y^2) < delta implies that |f(x,y)-L| < epsilon.

Normally I do this with something like: |f(x,y)-L| = .... = .... ≤ |y| or |x| ≤ sqrt(x^2 + y^2) < delta = epsilon.

Or I could show that there is no limit by using a two counterexamples (two functions on the curve sin(x-y)/cos(x+y) that approach different values in f(x,y) near (0,0)).

I found that on the line x=y, lim y->0 $sin(y-y)/cos(y+y) = sin(0)/cos(0) = 0$

But on the curve x = \pi*cos(y), lim y->0 $sin(\pi*cos(y)-y)/cos(\pi*cos(y)+y) = sin(\pi*cos(0)-0)/cos(\pi*cos(0)+0) = sin(\pi)/cos(\pi) = 1/0$, so there is no limit when approaching $sin(x-y)/cos(x+y)$ from the curve $x = \pi*cos(y).$

Doesn't it follow that $sin(x-y)/cos(x+y)$ has no limit at (0,0)? Or do I need to find a counter example of some x=f(y) where lim y->0 $sin(f(x)-y)/cos(f(x)+y)$ actually exists but is not zero?

2. ## Re: Finding & Proving the limit of lim (x,y)->(0,0) sin(x-y)/cos(x+y)

Hi,
Your text is right. In your 2nd curve, you said sin(pi)/cos(pi)=1/0, but sin(pi)/cos(pi)=0/(-1)=0.

3. ## Re: Finding & Proving the limit of lim (x,y)->(0,0) sin(x-y)/cos(x+y)

Also, for lim (x,y) -> (1,2) $(2x^2-xy)/(4x^4-y^2)$

on the line y = x, lim x->1 $(2x^2-x^2)/(4x^2-x^2) = 1/3$

on the line y=2x, the limit ends up with 0/0 (which I think suggests that I need to take a different approach to find the limit at y=2x)

on the line y = 3x, the limit is 1/5

So if I simplify $(2x^2-xy)/(4x^4-y^2)$ to just $x/(2x+y)$ by factoring out $(2x-y)$ from the top and bottom, now suddenly:

on the line y=x, the limit is 1/3 (same)

on the line y=2x, the limit is 1/4

on the line y = 3x, the limit is 1/5 (same)

So by approaching the curve from different directions I end up with different limits, but by simplifying my function to just $x/(2x+y)$, I can easily just substitute (x,y)=(1,2) no problem, and that should definitely be my limit. From what I learned though, the function has no limits if different lines on it approach different limits from each other?

4. ## Re: Finding & Proving the limit of lim (x,y)->(0,0) sin(x-y)/cos(x+y)

My bad johng, I guess what I had in mind was $x = (\pi/2)*cos(y)$, so that $sin(\pi/2)/cos(\pi/2) = 1/0$

5. ## Re: Finding & Proving the limit of lim (x,y)->(0,0) sin(x-y)/cos(x+y)

It seems like If I can directly sub in the values that x and y approach into f(x,y), maybe after factoring something out of the top and bottom of f(x,y), then that value is definitely the limit no matter what. When that is not the case, then the limit will be whatever it is that all curves on f(x,y) approach. If two of those curves have different limits on f(x,y) as x and y approach their limit points, then f(x,y) has no limit.
Does that sound about right?

6. ## Re: Finding & Proving the limit of lim (x,y)->(0,0) sin(x-y)/cos(x+y)

You should be able to use the properties of limits. If $f(x,y)$ is continuous at $(a,b)$, then $\lim_{(x,y)\to (a,b)}f(x,y) = f(a,b)$. For your second problem, $2x^4-y^2 = (2x^2+y)(2x^2-y)$ so you cannot factor out $2x-y$. For this problem, the function is not continuous at $(1,2)$ since $4(1)^4-(2)^2=0$. Now, you can check approaches. How about trying out lines that pass through (1,2)? $y=m(x-1)+2$ covers all non-vertical lines passing through (1,2). So, you have $\lim_{x \to 1} \dfrac{2x^2-x(m(x-1)+2)}{4x^4-(m(x-1)+2)^2}$. Multiplying out and simplifying, you have:

$\lim_{x\to 1} \dfrac{(m-2)x}{(m-2x-2)(2x^2+mx-m+2)} = \dfrac{m-2}{4(m-4)}$. Since you get different results for different values of $m$, the limit does not exist.

7. ## Re: Finding & Proving the limit of lim (x,y)->(0,0) sin(x-y)/cos(x+y)

Sorry but the $2x^4-y^2$ should be $2x^2-y^2$ so that it can be factored that way. I changed the exponent from 2 to 4 by mistake. What you say makes sense though. But basically my last comment "It seems life if I can directly sub..." is correct.

8. ## Re: Finding & Proving the limit of lim (x,y)->(0,0) sin(x-y)/cos(x+y)

Originally Posted by HowDoIMath
Sorry but the $2x^4-y^2$ should be $2x^2-y^2$ so that it can be factored that way. I changed the exponent from 2 to 4 by mistake. What you say makes sense though. But basically my last comment "It seems life if I can directly sub..." is correct.
You can "directly sub" if the function is defined at the limit point, but unless the function is continuous, the limit is not guaranteed to exist, much less equal the value of the function at the point. In other words, $f(a,b)$ may be defined, but $\lim_{(x,y) \to (a,b)} f(x,y)$ does not necessarily equal $f(a,b)$. In one dimension, this occurs when there is a removable discontinuity (requiring a function defined piecewise). In multiple dimensions, it is not nearly as obvious when $\lim_{(x,y) \to (a,b)} f(x,y) \neq f(a,b)$. For this problem, however, you can simplify the formula to a function that is continuous at the limit point.

$\dfrac{2x^2-xy}{4x^2-y^2}$ is a rational function, so it is continuous at all points where it is defined. So, given any $(x,y)$ such that $y\neq \pm 2x$, $\dfrac{2x^2-xy}{4x^2-y^2} = \dfrac{x}{2x+y}$ as you stated. Then, regardless of the choice of path, $\lim_{(x,y) \to (1,2)} \dfrac{2x^2-xy}{4x^2-y^2} = \lim_{(x,y) \to (1,2)} \dfrac{x}{2x+y}$. Since $\dfrac{x}{2x+y}$ is also a rational function, it is continuous at all values where $y \neq -2x$, so since $2 \neq -1$, it is continuous at $(1,2)$, allowing you to "directly sub in".