can anyone give me a hint on how to solve this probelem
The limit is equal to $\frac{\int_0^0 \sin(xt^3)dt}{0}= \frac{0}{0}$, so we can use De L'Hospital.We get:$$\lim_{x \to 0} \frac{\sin(x^4)}{5x^4}= \frac{1}{5} \lim_{x \to 0} \frac{\sin(x^4)}{x^4}= \frac{1}{5}$$
since it is known that $\lim_{u \to 0} \frac{\sin u}{u}=1$.
Hi everybody
If we have not any answers for critical points after first partial derivatives equal to zero, how can we continue to find local MAX, local MIN and Saddle point?. For example: Suppose we have below equations for first partial derivatives:
∂ƒ/∂x = y + 5 , ∂ƒ/∂y = 2z , ∂ƒ/∂z = y
As you can see, for ∇ƒ= 0 , there are not any answers (undefined)