MHB Solving Limits Using De L'Hospital's Rule

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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 $\frac{1}{5}$.
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can anyone give me a hint on how to solve this probelem View attachment 4167
 

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