Difficult calculus 2 question on using the fundamental theorem of calculus to evaluate limits I have about 5 questions of the same kind that I have no idea how to solve, the title of the work sheet is fundamental theorem of calculus.
the image of the question :
the questions
 A: For the first question, you are asked to evaluate
$$
\lim_{x\rightarrow 0}\frac{1}{\sin^2(x)}\int_0^x \frac{t}{\cos t}dt
$$
An integral that looks like $\int_a^af(t)dt$ evaluates to 0, because we are measuring the area under a function over a zero length interval. Then, since $\sin^2 (x)\rightarrow 0$ as $x\rightarrow 0$ we can see that this is an indeterminate form, and we can apply l'hospital's. 
Here is where the FTC comes in. Note that by the FTC
$$
\frac{d}{dx}\int_0^x \frac{t}{\cos t}dt=\frac{x}{\cos x}
$$
So by l'hospital's rule
$$
\lim_{x\rightarrow 0}\frac{1}{\sin^2(x)}\int_0^x \frac{t}{\cos t}dt=
\lim_{x\rightarrow 0}\frac{1}{2\sin(x)\cos x}\frac{x}{\cos x}=\\
\frac{1}{2}
\lim_{x\rightarrow 0}\frac{x}{\sin(x)\cos^2 x}
$$
And applying L'Hospital's again
$$
\frac{1}{2}
\lim_{x\rightarrow 0}\frac{x}{\sin(x)\cos^2 x}=
\frac{1}{2}
\lim_{x\rightarrow 0}\frac{1}{\cos(x)\cos^2 x-2\sin x \cos x}=1/2
$$
Since we have arrived at a continuous function through our work. The rest should be similar.
