GIF with Limits.

Calculus Level 4

Let $f(x)$ be a non-constant real valued polynomial function such that at the point $a$ we have $f(a)^2+f'(a)^2=0$ . Find the value of

$\lim_{x \to a} \dfrac{f(x)}{f'(x)} \cdot \left\lfloor \dfrac{f'(x)}{f(x)} \right\rfloor$

Note: $f'(x) = \dfrac{d}{dx} f(x)$

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Sandeep Bhardwaj
Apr 3, 2015

$\text{Given : }f(a)^2+f'(a)^2=0$

$\implies f(a)=f'(a)=0$

Clearly, $a$ is the repeated root of $f(x)=0$

$\therefore \lim_{x \to a} \dfrac{f(x)}{f'(x)} \cdot \left\lfloor \dfrac{f'(x)}{f(x)} \right\rfloor$

$\displaystyle =\lim_{x \to a} \dfrac{f(x)}{f'(x)} \cdot \left( \dfrac{f'(x)}{f(x)}- \displaystyle \left\{ \dfrac{f'(x)}{f(x)} \displaystyle \right\} \right)$

$\displaystyle =\lim_{x \to a} \left( 1-\dfrac{f'(x)}{f(x)} \cdot \left \{ \dfrac{f'(x)}{f(x)} \right \} \right)$

$=1-0=\boxed{1}$

Thanks. You should explain the 2nd last step a bit more. Note that there's a slight typo, where the first part of the second term should be $\frac{ f(x) } { f'(x) }$ .

The crucial part is that $\{ \frac{ f'(x) } { f(x) } \}$ tends to 0.

Calvin Lin Staff - 6 years, 2 months ago

I couldn't get how the fraction part limit is zero.

Navin Murarka - 3 years, 6 months ago

Can you please explain how to find the limit of your stated "crucial part" ??

Aaghaz Mahajan - 3 years, 4 months ago

$I\ took \ \ f(x)=x^2$ .

Rudresh Tomar - 5 years, 7 months ago

GIF with Limits.

Practice the set Target JEE_Advanced - 2015 and boost up your preparation.

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