Vtyjkyuk

I'm having a hard time evaluating the following limit

$$lim_x rightarrow 0 fracint_0^2 sin x cos(t^2) dt2x$$
I'm not even really sure how to approach it since I'm not used to seeing another function in the integral.

My initial thoughts were to use l'hopitals rule since $lim_x rightarrow0 2sin x=0$, so the numerator is equal to 0, but I wouldnt be sure how to take the derivative of the numerator.

Any pointers would be appreciated

We will use that $$f(x)=int_0^a(x)g(t)dtimplies f'(x)=g(a(x))a'(x).$$ In our case:

$$fracddx int_0^2 sin x cos(t^2) dt=cos((2sin x)^2)cdot dfracd(2sin x)dx.$$

So, we have

$$fracddx int_0^2 sin x cos(t^2) dt=2cos (x)cos(4sin^2 x).$$

Thus

$$lim_x rightarrow 0 fracint_0^2 sin x cos(t^2) dt2x=lim_x rightarrow 0dfrac2cos (x)cos(4sin^2 x)2 =1.$$

Without Liebnitz rule, you can use first mean value theorem for integrals, then there exist $c$ such that
$$lim_x rightarrow 0 fracint_0^2 sin x cos(t^2) dt2x=lim_x rightarrow 0 frac2 sin x cos(c^2) 2x=lim_x rightarrow 0 frac2 sin x 2xlim_x rightarrow 0 cos(c^2)=lim_x rightarrow 0 cos(c^2)=1$$
because with sandwich $0leq cleq2sin x$ so $cto0$ as $xto0$.

The denominator $2x$ can be replaced by $2sin x$ via the limit $lim_xto 0(sin x) /x=1$ and then putting $u=2sin x$ the limit is easily seen to be $$lim_uto 0 frac1uint_0^ucos(t^2),dt=cos(0^2)=1$$ via Fundamental Theorem of Calculus.

Alternatively, use Taylor expansion:
$$cos(t^2)=1-fract^42+fract^824+O(t^9);\
int_0^2sin x cos(t^2)dt=left(t-fract^510+O(t^6)right)bigg_0^2sin x=2sin x-frac(2sin x)^510+O((2sin x)^6);\
lim_x rightarrow 0 fracint_0^2 sin x cos(t^2) dt2x=lim_x rightarrow 0frac2sin x+O(sin^5x)2x=lim_x rightarrow 0frac2sin x2xcdot (1+O(sin^4x))=1.$$