# Use the method of your choice to evaluate the following limits. lim_{(x,y)rightarrow(0,1)}frac{ysin x}{x(y+1)}

Limits and continuity
Use the method of your choice to evaluate the following limits.
$$\lim_{(x,y)\rightarrow(0,1)}\frac{y\sin x}{x(y+1)}$$

2021-01-03
Given limits:
$$\lim_{(x,y)\rightarrow(0,1)}\frac{y\sin x}{x(y+1)}$$
We have to evaluate the limit.
The limit is computed as follows:
$$\lim_{(x,y)\rightarrow(0,1)}\frac{y\sin x}{x(y+1)}=\lim_{x\rightarrow0}\frac{\sin x}{x}\lim_{y\rightarrow1}\frac{y}{y+1}$$
Substitutes the limits:
We know that: $$\lim_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1$$
$$\lim_{(x,y)\rightarrow(0,1)}\frac{y\sin x}{x(y+1)}=1\times\frac{1}{1+1}$$
$$\lim_{(x,y)\rightarrow(0,1)}\frac{y\sin x}{x(y+1)}=\frac{1}{2}$$
Therefore, the value of $$\lim_{(x,y)\rightarrow(0,1)}\frac{y\sin x}{x(y+1)}$$ is $$\frac{1}{2}$$