The limit problem presented involves the expression $\frac{\int_0^0 \sin(xt^3)dt}{0}$, which evaluates to the indeterminate form $\frac{0}{0}$. To resolve this, De L'Hospital's Rule is applied, leading to the limit $\lim_{x \to 0} \frac{\sin(x^4)}{5x^4}$. This simplifies to $\frac{1}{5} \lim_{x \to 0} \frac{\sin(x^4)}{x^4}$, which is known to equal $\frac{1}{5}$ since $\lim_{u \to 0} \frac{\sin u}{u}=1$. The final result of the limit is $\frac{1}{5}$.
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 $\lim_{u \to 0} \frac{\sin u}{u}=1$.
Hello!
I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem.
Given:
##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0##
##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1##
##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0##
I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...