Solving Limits Using De L'Hospital's Rule

[bahadeen]

can anyone give me a hint on how to solve this probelem View attachment 4167

 

[evinda]

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 $\lim_{u \to 0} \frac{\sin u}{u}=1$.