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

Question
Limits and continuity
asked 2021-01-02
Use the method of your choice to evaluate the following limits.
\(\lim_{(x,y)\rightarrow(0,1)}\frac{y\sin x}{x(y+1)}\)

Answers (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 \(\frac{1}{2}\)
0

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