Can you find this limit or show that it doesn't exist?

Question

no i'm not necessarily just looking for the answer, but seriously how to do it and where to start. the problem is this:

limit (x,y)-->(0,0) (x^2+(siny)^2)/(2x^2+y^2)

thanks!

Answer 1

There is no limit.

You can see this by approaching from two different directions, along the x-axis and along the y-axis.If x is zero and y->0, then the formula becomes:

(siny)^2/y^2

Since sin y/y -> 1 as y->0, this also approaches 1 as y approaches zero.

On the other hand, if y=0, and x->0, then the formula becomes:

x^2/(2x^2)

which has a limit of 1/2 as x->0.

Answer 2

First decide whether you think the limit exists or not. In this case, we know that sin y is about the same as y if y is close to 0 since
lim sin(y)/y=1 in this limit.
This means that we have essentially
(x^2+y^2)/(2x^2+y^2)
with a 0/0 limit.
The rule of thumb is that limits of polynomials of the same degree will not exists, so you have to find two paths to (0,0) that give different answers. Typically, for the limit to exist, the polytnomial on top has to be of larger degree.
Now, looking at (x,0) as x->0 we get a limit of 1/2.
Looking at (0,y) as y->0, we get a limit of 1.
This difference confirms that the limit fails to exist.

Answer 3

dude,first f ol i might lyk 2 tell u that shrink will exist on condition that L.H.L=R.H.L define the functionality, |h| { h if h>=0, {-h if h<0 , for L.H.L limh»(0-) -h/h =-a million, for R.H.L limh»(0+) h/h =a million, so shrink doesnt exist, 2) suggesting u one undertaking,if u wil study graphs of the multiple uncomplicated applications like properly suited integer and modulus functionality,u can handle the pbms. lot much less complicated..thank you,n sturdy success.

Answer 4

Danger Will Robinson.

I earned a C- in Beginning Algebra,but I passed at Los Angeles Valley College.

Dr. Donald Mazukelli was the professor who was strict. His motto was "I do not want robots."

Dr. Glen Thomas who was my Engineering professor also retired and I think it was about 3 years ago.

Answer 5

Use L'Hospital's theorem.