91Ó°ÊÓ

As we can see, the given expression is a two-variable limit. This means that we have to make sure the limit is consistent as we approach (0,1) from different paths. The expression is given by: $$\frac{y \sin x}{x(y+1)}$$ We can notice that if we try to directly substitute the values x=0 and y=1, we get an indeterminate form \(\frac{0}{0}\). Therefore, a different approach is needed.

Let's make a substitution to simplify the expression. We can substitute \(u = y + 1\). This means that \(y = u - 1\). Now, we rewrite the expression in terms of u: $$\frac{(u-1) \sin x}{x(u)}$$ Our new limit is: $$\lim _{(x, u) \rightarrow(0,2)} \frac{(u-1) \sin x}{x(u)}$$

Similar to step 1, when we try to substitute the values x=0 and u=2, we still get an indeterminate form \(\frac{0}{0}\). Therefore, we should consider approaching the limit from different paths to see if the limit is consistent. Try to find the limit along the lines x=0 and u=2. Along the line x=0, the expression becomes: $$\frac{(u-1) \sin 0}{0(u)} = 0$$ Along the line u=2, the expression becomes: $$\frac{(\sin x)}{x(2)}$$ We can use the Squeeze theorem to find the limit above as x approaches 0: $$\lim_{x \to 0} {-\frac{1}{2}} \leq \lim_{x \to 0} {\frac{\sin x}{x}} \leq \lim_{x \to 0} {\frac{1}{2}}$$ Therefore \(\lim_{x \to 0} {\frac{\sin x}{x}} = \frac{1}{2}\), which implies that as the point (x,u) approaches (0,2) along u=2, the limit is given by: $$\lim_ {(x, u) \rightarrow (0,2)} \frac{(\sin x)}{x(2)} = \frac{1}{4}$$

Since the limit along the line x=0 is 0, and the limit along the line u=2 is \(\frac{1}{4}\), it is clear that the limit will not be consistent as the point (x, y) approaches (0,1) because different paths give different results of the limit. Therefore, we conclude that the given limit does not exist: $$\lim _{(x, y) \rightarrow(0,1)} \frac{y \sin x}{x(y+1)}$$