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Differential Calculus

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15 Jul 2008 13:54:01 IST

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798

a challenging prob for all goiitians

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Suppose be a real valued function defined for all satisfying and

f ' (x) = 1/[x² + f(x)²]

Prove that the limit of f(x) as x goes to infinity exists and is less that

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Hari Shankar

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15 Jul 2008 14:08:58 IST

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Nice prob!

I am tempted! Can I atleast give a vague hint to the students?

Anant Kumar

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15 Jul 2008 14:12:42 IST

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sure ... go ahead

Hari Shankar

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15 Jul 2008 14:15:21 IST

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This problem should have been posted in the Integral Calculus section!!

Vague enough sir?

Anant Kumar

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15 Jul 2008 14:18:35 IST

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quite a hint :)

Hari Shankar

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15 Jul 2008 19:00:48 IST

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ummm anyone trying? I am waiting till 9 am. 2moro before posting. Site ka izzat ka sawaal hey!

Hari Shankar

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16 Jul 2008 09:41:52 IST

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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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