How can I prove $\lim_{(x,y)\rightarrow(3,5)}\sqrt[3]{x+y}=2$ by definition?

Question

$$\lim_{(x,y)\rightarrow(3,5)}\sqrt[3]{x+y}=2$$

The cubic root over there makes this problem a little tricky for me, i don't really know how to proceed. Any ideas?

Answer 1

Proving this limit equation directly from the $\epsilon$,$\delta$ definition is possible, but tricky. It requires you to somehow deal with the inequality $$\bigl| \sqrt[3]{x+y}-2 \bigr| < \epsilon $$

Have you ever seen what to do when you have a square root inequality instead of a cube root? For example, let's consider the simpler problem of proving that $\lim_{(x,y) \to (1,3)} \sqrt{x+y}=2$ then you would have to deal with the inequality $$\bigl| \sqrt{x+y}-2 \bigr| < \epsilon $$ You can work with this inequality using the trick of "rationalizing the numerator", which is based on the formula $(a-b)(a+b)=a^2-b^2$ i.e. $$|\sqrt{x+y}-2| = \left| \frac{(\sqrt{x+y}-2)(\sqrt{x+y}+2)}{\sqrt{x+y}+2} \right| = \left| \frac{x + y - 4}{\sqrt{x+y}+2} \right| $$ Then you can introduce appropriate lower bounds on the denominator: assuming, for example, that $|x+y-4| < 1$ it follows that $x+y > 3$ and that $\sqrt{x+y} + 2 > \sqrt{3}+1$ and so $$\left| \frac{x + y - 4}{\sqrt{x+y}+2} \right| \le \frac{1}{\sqrt{3}+2} |x+y-4| $$ And, finally, you let $\delta = \min\{1,(\sqrt{3}+2)\epsilon\}$ and continue on with the proof.

So now, since what you actually have is a cube root inequality, you can use a similar "rationalizing the numerator" trick, based on the formula $(a-b)(a^2+ab+b^2)=a^3-b^3$. It's messy, but I'll write out the outcome with the middle step skipped: $$|\sqrt[3]{x+y}-2| = \left| \frac{x+y-8}{(\sqrt[3]{x+y})^2 + 2 \cdot \sqrt[3]{x+y} + 4} \right| $$ And again, since it's only addition in the denominator, now you can introduce appropriate lower bounds on the denominator: if, say $|x+y-8|<1$ then $x+y > 7$ and that gives you a positive lower bound on the denominator which you can continue with, something like this (if I've done the arithmetic correctly): $$(\sqrt[3]{x+y})^2 + 2 \cdot \sqrt[3]{x+y} + 4 > 7^{2/3} + 2 \cdot 7^{1/3} + 2 $$ and so you'll let $\delta = \min\{1, (7^{2/3} + 2 \cdot 7^{1/3} + 2) \epsilon\}$ and continue on with the proof.

Answer 2

HINT: $f(x,y)$ is differentiable at $(a,b)$ if $(a,b)$ , $f_x(a,b)$,$f_y(a,b)$ exists and $\lim_{x,y \to a,b}\frac{f(x,y)-L(x,y)}{\sqrt{(x-a)^2 + (y-b)^2}}=0$ where $L(x,y)=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$

L is called linear approximation..