Multivariable delta-epsilon limit proof

Question

Prove using the delta epsilon definition that

$$ L = \lim_{(x,y)\rightarrow (0,0)}\frac{x^5e^{xy}}{x^2+e^y}=0 $$

What I have done so far: $$\left | \frac{x^5e^{xy}}{x^2+e^y}-0 \right |< \epsilon \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0<\sqrt{x^2+y^2}<\delta$$

Then $$ \left | \frac{x^5e^{xy}}{x^2+e^y}-0 \right |=\left | \frac{x^5e^{xy}}{x^2+e^y} \right |\leq \left | \frac{x^5e^{xy}}{x^2} \right |=\left | x^3 \right |\left | e^{xy} \right |=... $$

And this is where I get stuck, how should I proceed?

Answer 1

Note that for $|x|,|y|<1$ you have that $|e^{xy}|<e$ and $|x^3|\leq|x^2|\leq|x^2+y^2|$.

Can you proceed, choosing $\delta$ small enough so that the given expression is less than $\epsilon$?

Answer 2

Be careful! You will also want to take limit along $x=0$ too so you shouldn't leave $x^2$ alone in the denominator.

Instead, look at $$ \left\lvert\frac{x^5 e^{xy}}{x^2+e^y}\right\rvert \leq \frac{\lvert x^5 e^{xy}\rvert}{e^y}=\lvert x^5\rvert\cdot e^{(x-1)y} $$ where the inequality comes from $e^y>0$ and $x^2\geq 0$.

Answer 3

$$ L = \lim_{(x,y)\rightarrow (0,0)}\frac{x^5e^{xy}}{x^2+e^y}=0 $$

Now $$\left | \frac{x^5e^{xy}}{x^2+e^y}-0 \right |< \epsilon \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 0<\sqrt{x^2+y^2}<\delta$$

Then $$ \left | \frac{x^5e^{xy}}{x^2+e^y}-0 \right |=\left | \frac{x^5e^{xy}}{x^2+e^y} \right |\leq \left | \frac{x^5e^{xy}}{e^y} \right |=\left | x^5 \right |e^{(x-1)y} $$

$$ <\left | x^5 \right |e\leq \left | x^2 \right |e\leq \left | x^2+y^2 \right |e\leq \left | x+y \right |e $$ $$ \leq e\sqrt{x^2+y^2}<e\delta $$ Thus

$$ e\delta=\epsilon \Rightarrow \delta=\frac{\epsilon}{e} $$