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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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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