We want to find the limit if it exists.
We have the limit of x comma y approaches two comma one
of the quantity x minus two y divided by
the quantity x minus two y squared.
So the first thing we should recognize here is
that we cannot find this limit performing
direct substitution because of the point two comma one
we would have zero divided by zero,
which is indeterminate, which means to find this limit
or see if it exists, we'll have to determine the limit
along several paths that approach the point two comma one.
So notice how on the right here I've graphed the point
two comma one on the x y coordinate plane,
and I've also graphed several paths that we could use
that approach this point.
To start with, let's use the two most obvious paths,
which would be y equals one and x equals two.
Now if we find these two limits along these paths,
and they're not equal, we can conclude
that the limit does not exist.
If they are equal, that does not tell us
the limit does exist, that means we'd have
to consider additional paths before we conclude
the limit exists and what that limit would be.
However it is often helpful to graft the function
and analyze the surface when determining limits
of functions of two variables.
And I've provided the graph of our surface here below.
Notice how the graph of the surface looks strange
along this path here,
which if we take a look at our function,
and recognize that we'd have discontinuity,
when we have division by zero,
so we set the denominator equal to zero,
so we set x minus two y squared equal to zero
and solve for x by adding two y squared to both sides
we'd have x equals two y squared.
So graphically what we're seeing here,
are these are the points where we have division by zero
and therefore the surface has discontinuity.
So this is something that can be helpful if we notice this.
But again what we're going to do now is consider the limit
along the path y equals one and x equals two.
So for the path of y equals one,
we'll substitute one for y, so we'd have the limit
of x comma one approaches two comma one of,
again if y equals one, our numerator would be
x minus two times one or just x minus two,
our denominator would be x minus two times one squared,
which is also just x minus two.
Notice here we can simplify this rational expression
x minus two divided by x minus two equals one
so this simplifies to the limit
of x comma one approaches two comma one of just one,
which of course just equals one.
So we've shown along the path y equals one
the limit would be positive one,
this does not mean this limit equals one
or that it even exists, we must consider additional paths.
Now let's consider the path along
this vertical line x equals two.
So for the path x equals two,
we'll now substitute two for x,
so we'd have the limit of two comma y
approaches two comma one of...
So if we substitute two for x, notice how we'd have
two minus two y divided by two minus two y squared.
And this is going to simplify,
let's do some algebra down here.
We factor out two from the numerator,
we'd have two times the quantity one minus y,
and in our denominator we'll factor out two as well,
so we have two times the quantity one minus y squared.
Notice how our denominator is a difference of squares.
So we have two times the quantity of one minus y,
divided by two, and then we have a difference of squares.
So we'd have the quantity one minus y
times the quantity one plus y.
And now when we simplify, notice how we have a common factor
of two, as well as common factor of y minus one,
so this simplifies to one divided by
the quantity one plus y.
Which means, now we have the limit
of two comma y approaches two comma one
of just one divided by the quantity one plus y.
So notice as y approaches positive one,
our limit would be one divided by two or 1/2.
And we can actually stop here,
because notice how we did not get the same limit,
the limit along the path of x equals two is 1/2,
the limit along y equals one is one,
and for this limit to exist,
the limit along all paths approaching two comma one
must be the same and since they're not,
we know this limit does not exist.
Okay, hope you found this helpful.