# Proving that multivariable limit exists

1. Aug 18, 2015

### physdood

1. The problem statement, all variables and given/known data
Given f(x,y)=(y+x)/(y-x) use an ε-∂ proof to show that lim(x,y)→(0,1) f(x,y) exists.

2. Relevant equations
|(y+x)/(y-x)-1|=|(2x)/(y-x)|

3. The attempt at a solution
I know that the limit is 1. I cant 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.

2. Aug 18, 2015

### Ray Vickson

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.

3. Aug 18, 2015

### 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
$d\big((x,y),(0,1)\big)<\delta\Rightarrow |x-0|<\alpha\wedge|y-1|<\alpha$ .

4. Aug 19, 2015

### Zondrina

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 $\delta( \varepsilon)$.