How to evaluate a limit that contains an integral? This question is from the JEE Advanced 2007 Paper 1 (Question 49). The question states:
$$\lim \limits_{x \to \frac{\pi}{4}} \frac{\int^{\sec^2x}_2 f(t)\ dt}{x^2-\frac{\pi^2}{16}}$$
I tried solving it using the following steps:
Step 1: $$\lim \limits_{x \to \frac{\pi}{4}} \frac{f(\sec^2x)-f(2)}{x^2-\frac{\pi^2}{16}}$$
Step 2: Substitute the value of the limit
$$\lim \limits_{x \to \frac{\pi}{4}} \frac{f(2)-f(2)}{\frac{\pi^2}{16}-\frac{\pi^2}{16}}$$
which gives $\frac00$
I couldn't solve further because I didn't know how to, and moreover $f\ $ is not given and there is no answer available on the internet.
 A: Alright so we want to compute the limit
$$L=\lim_{x\to\frac{\pi}{4}}\frac{\int_2^{\sec^2 x}f(t)~\mathrm{d}t}{x^2-\frac{\pi^2}{16}}.$$
Since no information is given about $f$, I'll just assume it to be "nice enough" for what I'm about to do (continuity should be enough). Now as you noticed, we have a $\frac{0}{0}$ indeterminate form, and so we can apply L'Hôpital's rule to solve it. Recall that
$$\frac{\mathrm{d}}{\mathrm{d}x}\sec^2 x=2\tan x\sec^2 x,$$
which gives us (using L'Hôpital's rule along with the FTC and the chain rule) that
$$L=\lim_{x\to\frac{\pi}{4}}\frac{2\tan x\sec^2 x f(\sec^2 x)}{2x}.$$
This we can evaluate by just plugging in $x=\frac{\pi}{4}$, which yields that
$$L=\frac{\tan\frac{\pi}{4}\sec^2\frac{\pi}{4}f\left(\sec^2\frac{\pi}{4}\right)}{\frac{\pi}{4}}=\frac{8}{\pi}f(2).$$
A: The key thing to note here is that as $x\to \pi/4$ we have $\sec^2x\to 2$ and hence we can rewrite the expression under limit as $$\frac{1}{\sec^2x-2}\int_2^{\sec^2x}f(t)\,dt\cdot\frac{\sec^2x-2}{x^2-(\pi/4)^2}$$ The first factor tends to $f(2)$ via fundamental theorem of calculus and the second factor can be written as $$\frac{\sec x-\sec(\pi/4)}{x-\pi/4}\cdot \frac {\sec x +\sqrt {2}}{x+\pi/4}$$ The second factor tends to $4\sqrt {2}/\pi$ and first factor tends to $\sqrt{2}$ (try it out). The desired limit is thus $8f(2)/\pi$.
The above approach avoids the use of LHospital Rule.
A: Hint: By l'Hôpital's rule and the fundamental theorem of calculus,
$$\lim_{x\to\frac\pi4} \frac{\int_2^{\sec^2(x)} f(t) \, dt}{x^2 - \frac{\pi^2}{16}} = \lim_{x\to\frac\pi4} \frac{(\sec^2(x))' f(\sec^2(x))}{2x}$$
Can you finish from here?
