prove using $\epsilon-\delta$ method that $\lim \limits _{(x,y)\rightarrow (0,0)}\frac{e^y\sin x}{x}=1$

Question

I have $\frac{|e^y\sin x-x|}{|x|}\leq\frac{|e^y-x|}{|x|}\leq\frac{e^y}{\sqrt{x^2}}$

I am unable to proceed and find $\delta$

Answer 1

Hint: using $$ \bigg|\frac{\sin x}{x}\bigg|\le1$$ one has $$ \bigg|\frac{e^y\sin x}{x}-1\bigg|=\bigg|\frac{(e^y-1)\sin x+\sin x-x}{x}\bigg|\le|e^y-1|+\bigg|\frac{\sin x}{x}-1\bigg|.$$ Then using this for $$\lim_{x\to0} \frac{\sin x}{x}=1$$ one will get $\delta$ for $\epsilon>0$.