Proving that multivariable limit exists

[physdood]

Homework Statement

Given f(x,y)=(y+x)/(y-x) use an ε-∂ proof to show that lim_{(x,y)→(0,1)} f(x,y) exists.

Homework Equations

|(y+x)/(y-x)-1|=|(2x)/(y-x)|

The Attempt at a Solution

I know that the limit is 1. I can't figure out how to massage the above any further to get it into the form |(2x)/(y-x)|<=k|x| or |(2x)/(y-x)|<=k|y-1| so that I can choose an appropriate value for ∂. I have tried restricting ∂<1 but it doesn't get me any further. Any hints would be appreciated.

Thanks.

 

[Ray Vickson]

Homework Statement

Given f(x,y)=(y+x)/(y-x) use an ε-∂ proof to show that lim_{(x,y)→(0,1)} f(x,y) exists.

Homework Equations

|(y+x)/(y-x)-1|=|(2x)/(y-x)|

The Attempt at a Solution

I know that the limit is 1. I can't figure out how to massage the above any further to get it into the form |(2x)/(y-x)|<=k|x| or |(2x)/(y-x)|<=k|y-1| so that I can choose an appropriate value for ∂. I have tried restricting ∂<1 but it doesn't get me any further. Any hints would be appreciated.

Thanks.

In problems of this type is often useful to translate the desired point to the origin. By writing $x = x'$ and $y = 1 + y'$, the limit becomes $(x',y') \to (0,0)$. Try it and see what happens.

 

[andrewkirk]

Saying the limit exists and equals 1 is saying that $\forall\epsilon>0\ \exists\delta>0$ such that $d\big((x,y),(0,1)\big)<\delta\Rightarrow |f(x,y)-1|<\epsilon$, where $d\big(x1,y1),x2,y2)\big)\equiv\sqrt{(x1-x2)^2+(y1-y2)^2}$ is the distance function.

So what you need to do is show how to find a $\delta$ that works, given $\epsilon$.

The problem is easier to do if you concentrate on a square in the number plane centred on (0,1), rather than the circle defined by the distance formula. So, given $\epsilon$, try to find $\alpha>0$, defined in terms of epsilon, such that if $|x-0|<\alpha$ and $|y-1|<\alpha$ then $|f(x,y)-1|<\epsilon$. That will be a matter of simple arithmetic using the formulas you have written above.

If you can do that, you only need to find $\delta$ such that the circle around (0,1) defined by $\delta$ contains the square around (0,1) defined by $\alpha$ so that
$d\big((x,y),(0,1)\big)<\delta\Rightarrow |x-0|<\alpha\wedge|y-1|<\alpha$ .

 

[STEMucator]

You got to the point:

$$\frac{2 |x|}{|y - x|} < \varepsilon$$

Using the reverse triangle inequality, you get:

$$\frac{2 |x|}{|y - x|} \leq \frac{2 |x|}{\left| |y| - |x| \right|} \leq \frac{2 |x|}{|y| - |x|} < \varepsilon$$

You know $|x| < \delta$ and $|y - 1| < \delta$. Use these inequalities to put a bound on $\frac{1}{|y| - |x|}$ in terms of $\delta$. This will let you solve for $\delta( \varepsilon)$.