# Evaluate $\lim_{(x,y)\to (0,0)}\frac{x+\sin y}{x+y}$

Evaluate if it exists $$\lim_{(x,y)\to (0,0)}\frac{x+\sin y}{x+y}$$

The answer in the book uses the fact the the iterated limits are equal to $1$ and if we take $x=-\sin y$ the limit is $0$ and then by theorem the limit does not exists

Is there a different way to show the limit does not exist?

Take $\displaystyle (x,y)=\left(\frac{1}{n},\frac{1}{n}\right)$ Then $$f\left(\frac{1}{n},\frac{1}{n}\right)=\frac{n}{2}\left(\frac{1}{n}+\sin\left(\frac{1}{n}\right)\right)=\frac{1}{2}+\frac{n}{2}\sin\left(\frac{1}{n}\right)\underset{n \rightarrow +\infty}{\rightarrow}1$$ Now for $\displaystyle x=-\sin\left(\frac{1}{n}\right)$ and $\displaystyle y=\frac{1}{n}$ you have that the limit is $0$.

Hence it does not have a limit.

• $lim_{n\to \infty} \frac{2sin(\frac{1}{n})}{sin(\frac{1}{n})+\frac{1}{n}}=1$ – gbox Mar 5 '18 at 21:48
• Just forgot a little $-$, i edit – Atmos Mar 5 '18 at 22:09

1) $y=0, x\not=0:$

$\lim_{(x,y) \rightarrow (0,0)} \dfrac{x+\sin y}{x+y}=$

$\lim_{(x,0) \rightarrow (0,0)}\dfrac{x}{x} =1$.

2) $x = -\sin y$ , as suggested.