Find $\lim_{x\to 0}(1+\int_{2x}^{4x}\sin(t^2) dt)^{\csc(4x^3)}$

Question

Find the exact value of $\lim_{x\to 0}(1+\int_{2x}^{4x}\sin(t^2) dt)^{\csc(4x^3)}$. I guess I need to $\log$ it, so I guess the limit is $\exp(\lim_{x\to 0}\frac{\ln(1+\int_{2x}^{4x}\sin(t^2)d t)}{\sin (4x^3)})$. However, L'opital rule does not help me here. How should I tackle it from this step?

Answer 1

Hint: $$ \lim_{x\to 0} \frac{\ln \left(1+\int_{2x}^{4x} \sin(t^{2})dt\right)}{\sin (4x^{3})} = \frac{1}{4} \lim_{x\to 0} \frac{4x^{3}}{\sin(4x^{3})}\frac{\ln \left(1+\int_{2x}^{4x} \sin(t^{2})dt\right)}{\int_{2x}^{4x}\sin(t^{2})dt}\frac{\int_{2x}^{4x}\sin(t^{2})dt}{x^{3}}. $$ The first and second limit is 1, and the last one $$ \lim_{x\to 0} \frac{\int_{2x}^{4x}\sin(t^{2})dt}{x^{3}} = \frac{56}{3} $$ can be proved by using l'Hospital's rule or by Taylor's theorem.