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 \(\frac{1}{2}\)

\(\lim_{(x,y)\rightarrow(0,1)}\frac{y\sin x}{x(y+1)}\)

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 \(\frac{1}{2}\)