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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.

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    $\begingroup$ Apply L'Hôpital's rule. Also, you should use $F(t)$ for the antiderivative function in step 1 rather than $f(t)$ which is your integrand. $\endgroup$
    – KStarGamer
    Aug 11, 2022 at 15:45
  • $\begingroup$ Have you tried applying Leibniz's integral rule? en.m.wikipedia.org/wiki/Leibniz_integral_rule $\endgroup$
    – Gerald
    Aug 11, 2022 at 15:51
  • $\begingroup$ If you factor the denominator, you can also view part of this limit as the derivative of $F(x)=\int_2^{\sec^2(x)}f(t) \, dt$. $\endgroup$
    – Sambo
    Aug 11, 2022 at 15:57

3 Answers 3

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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

$$L=\frac{\tan\frac{\pi}{4}\sec^2\frac{\pi}{4}f\left(\sec^2\frac{\pi}{4}\right)}{\frac{\pi}{4}}=\frac{8}{\pi}f(2).$$

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  • $\begingroup$ If I may ask, what do you mean by "FTC"? $\endgroup$ Aug 11, 2022 at 17:58
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    $\begingroup$ @integralmaths Fundamental Theorem of Calculus (context should make clear which part). $\endgroup$
    – J.G.
    Aug 11, 2022 at 18:14
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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.

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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?

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  • $\begingroup$ Note: this also uses the chain rule (when applying the FTC). $\endgroup$
    – Sambo
    Aug 11, 2022 at 15:55

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