91Ó°ÊÓ

Observe the given expression that we need to compute the limit for: $$\lim _{(x, y) \rightarrow(1,0)} \frac{y \ln y}{x}$$

To analyze the limit of a two-variable function approaching a point, it's often helpful to change the coordinates to polar form. Let's initiate the coordinate transformation by setting x = r*cos(θ) and y = r*sin(θ). The limit expression becomes: $$\lim _{r \rightarrow 0} \frac{r\sin \theta \ln (r\sin \theta)}{r\cos\theta}.$$

Now, we can simplify the expression by canceling out the 'r' term in both the numerator and denominator, which gives us: $$\lim _{r \rightarrow 0} \frac{\sin \theta \ln (r\sin \theta)}{\cos\theta}.$$

As r approaches 0, the term 'r*sin(θ)' in the logarithm will also approach 0. However, since the natural logarithm of 0 is not defined, we cannot directly substitute the limit. But we can use L'Hôpital's rule in this case. L'Hôpital's rule states that if the limit of the ratio of two functions is indeterminate (0/0 or ∞/∞), then the limit of their derivatives is the same as the limit of the ratio of the original functions. So, to use L'Hôpital's rule in our case: $$\lim _{r \rightarrow 0} \frac{\sin \theta \ln (r\sin \theta)}{\cos\theta} = \lim _{r \rightarrow 0} \frac{\frac{\partial}{\partial r}(\sin \theta \ln (r\sin \theta))}{\frac{\partial}{\partial r}(\cos\theta)}.$$ Now, since \(\cos\theta\) doesn't depend on r, its partial derivative with respect to r is 0. For the partial derivative of the numerator (product rule): $$\frac{\partial}{\partial r}(\sin \theta \ln (r\sin \theta)) = \sin\theta \cdot \frac{1}{r\sin\theta} \cdot \sin\theta + \ln(r\sin\theta) \cdot 0 = \frac{\sin^2\theta}{r\sin\theta}.$$ Now the limit expression becomes: $$\lim _{r \rightarrow 0} \frac{\frac{\sin^2\theta}{r\sin\theta}}{0}.$$

We observe that the limit expression still remains undefined, and the limit cannot be determined for all values of θ. As r approaches 0, the expression \(\frac{\sin^2\theta}{r\sin\theta}\) will fluctuate between positive and negative values, depending on the value of θ. Therefore, we can conclude that: $$\lim _{(x, y) \rightarrow(1,0)} \frac{y \ln y}{x}$$ does not exist.