Polynomial degrees are back again

Calculus Level 5

$\large \lim_{t \to a} \frac {\int_a^t f(x) \ dx - \frac {t-a}{2} \left (f(t) + f(a) \right ) }{(t-a)^3} = 0$

If $f(x)$ is a polynomial that satisfy the limit above for all $a$ , then the degree of $f(x)$ can at most be?

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1 solution

Harshvardhan Mehta
Jan 26, 2015

We have, $\underset { t\rightarrow a }{ lim } \frac { \int _{ a }^{ t }{ f(x)dx-\frac { (t-a) }{ 2 } (f(t)+f(a)) } }{ { (t-a) }^{ 3 } }=0$

Applying L' Hospitals's rule,

$\Rightarrow \underset {t\rightarrow a}{lim}\frac{f(t)-\frac{1}{2}(f(t)+f(a))-\frac{(t-a)}{2}f'(t)}{3(t-a)^2}=0$ $\Rightarrow \underset {t\rightarrow a}{lim}\frac{f'(t)-\frac{1}{2}(f(t)+f(a))-\frac{(t-a)}{2}f'(t)}{12(t-a)}=0$ $\Rightarrow \frac{f''(t)}{12}=0$

$f''(a)=0 \ for \ any \ a.$

$\Rightarrow f(a) \ is \ atmost \ of \ degree \boxed{1}.$

Moderator note:

Jon Haussmann is right. Why do you have to differentiate $a$ too? Note that we are only apply L'hopital on the limit itself, not differentiating EVERYTHING with respect to $t$ .

Why is the limit equal to $a$ in the first equation, then equal to 0 in the second equation?

Jon Haussmann - 6 years, 4 months ago

it is equal to a only please check carefully.

Harshvardhan Mehta - 6 years, 4 months ago

You go from $\underset { t\rightarrow a }{\lim } \frac { \int _{ a }^{ t }{ f(x)dx-\frac { (t-a) }{ 2 } (f(t)+f(a)) } }{ { (t-a) }^{ 3 } }=a$ to $\underset {t\rightarrow a}{\lim}\frac{f(t)-\frac{1}{2}(f(t)+f(a))-\frac{(t-a)}{2}f'(t)}{3(t-a)^2}=0$ by applying L'Hopital's rule, so the two limits are the same. So why is the first limit equal to $a$ , and the second limit equal to 0?

By the way, the LaTeX code for limit is \lim.

Jon Haussmann - 6 years, 4 months ago

@Jon Haussmann – @Harshvardhan Mehta

Keshav Tiwari - 6 years, 2 months ago

@Jon Haussmann – Well just checked it now and realized that it was a typing error thanks for bringing to my notice. $\ddot \smile$

Harshvardhan Mehta - 6 years, 1 month ago

You made a mistake in the third equation. It should be: $\lim_{t \to a} \dfrac{f(t)-f(a) - (t-a)f'(t)}{6(t-a)^2}$

Ariel Gershon - 4 years, 7 months ago

This question has already appeared in JEE

Shubhendra Singh - 4 years, 7 months ago

This should be reported as the ans is it should be of a second degree,if it is of a single one,the numerator will be of a second degree and hence it cannot converge to a finite limit there

Chirag Shyamsundar - 4 years, 7 months ago

Polynomial degrees are back again

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