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Discussion Response Post to: a challenging prob for all goiitians

Forum Index -> Differential Calculus -> View Full Question

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Hari Shankar (9109)

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total posts: 2173
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We are given that

This means f'(x)>0 for such x.

i.e. f(x)>1 if x>1

Now we have

f(x) - f(1) = $\int_1^{x} f$

Hence,

$\lim_{x \rightarrow \infty} (f(x) - f(1)) = \int_1^{\infty} f$

$\int_1^{\infty} \frac{1}{1+x^2} dx = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \\ \\<br/>\therefore \lim_{x \rightarrow \infty} f(x) < f(1) + \frac{\pi}{4} = 1+\frac{\pi}{4}$

Thus f is a monotonically increasing function bounded above and hence the limit exists and is less than $1+\frac{\pi}{4}$

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