$$f(x)= \begin{cases} \dfrac{\sin(xy)}{\sqrt{(x^4+y^4)}} & \text{if } (x,y)\ne(0,0), \\[10pt] 0 & \text{otherwise.} \end{cases}$$
Compute limit at $(0,0)$ along the line $y=mx$
I substitute in $y=mx$
$$\lim_{(x,y)\to(0,0)} f(x)= \lim_{(x,y)\to(0,0)} \frac{\sin(mx^2)}{\sqrt{(x^4+(mx)^4)}}$$
I can see that the numerator goes to $0$ and the denominator goes to $\infty$. But I'm unsure how to proceed and show that the limit doesn't exist.
Thanks
Best Answer
$$\lim_{(x,y)\to(0,0)}\dfrac{\sin mx^2}{\sqrt{x^4+(mx)^4}} = \lim_{(x,y)\to(0,0)}\dfrac{\sin mx^2}{mx^2}.\dfrac{mx^2}{x^2\sqrt{1+m^4}}=\dfrac{m}{\sqrt{1+m^4}}$$