# Limits using derivatives

If a differentiable function $f(x)$ satisfies a functional rule $f(x) + f(x+2) + f(x+4) = 0$ for all $x$ belonging to real numbers,

Then find the value of :

$$\lim \limits_{x \to 0} \frac{\bigl(f(x+12)\bigr)^2- f(x) f(0) - f(x+6) f(18) + \bigl(f(18)\bigr)^2}{x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)}$$

what's the method?

The answer is $32$.

• If the limit goes to 1 the limit is just $(\text{something})/\frac\pi4$. Are you sure the equation is copied correctly? Also, $f(x)=0$ is a differentiable function satisfying the functional equation but the limit is obviously 0 not 32. – kennytm May 15 '16 at 14:11
• There is obviously a mistake in question (perhaps the book has typo). Because from the given hypotheses we can't conclude that limit is $32$. See the answer by user A s. Also the comment by @kennytm is valid. – Paramanand Singh May 16 '16 at 2:56
• I have corrected the question it is limit goes to 0 not 1. Please note. – Harry Karwasra May 16 '16 at 8:11
• Even if you change $x \to 1$ to $x \to 0$ the answer is not guaranteed to be $32$ rather the answer is equal to $2\{f'(0)\}^{2}$. Also as noted by @kennytm the function $f(x) = 0$ satisfies all the requirements of the question and then the limit is $0$. – Paramanand Singh May 16 '16 at 9:53

If dont know if the following can help you \begin{align} &f(x) + f(x+2) + f(x+4) = 0\hspace{1cm} (I)\\ &f(x+2) + f(x+4) + f(x+6) = 0\hspace{1cm} (II)\\ &f(x+6) + f(x+8) + f(x+10) = 0\hspace{1cm} (III)\\ &f(x+8) + f(x+10) + f(x+12) = 0\hspace{1cm} (IV)\\ &f(x+12) + f(x+14) + f(x+16) = 0\hspace{1cm} (V)\\ &f(x+14) + f(x+16) + f(x+18) = 0\hspace{1cm} (VI)\\ \end{align} First $(I)-(II)$ gives $f(x)=f(x+6)$ In particular $f(0)=f(6)$
also $(I)-(II)+(III)-(IV)$ gives $f(x)=f(x+12)$ In particular $f(0)=f(12)$
also also $(I)-(II)+(III)-(IV)+(V)-(VI)$ gives $f(x)=f(x+18)$
etc ... etc .. Noting that $\lim \limits_{x \to 1} x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)=\frac{\pi}{4}$. So $$\lim \limits_{x \to 1} \frac{\bigl(f(x+12)\bigr)^2- f(x) f(0) - f(x+6) f(18) + \bigl(f(18)\bigr)^2}{x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)}= \frac{4}{\pi} \lim \limits_{x \to 1} (f(x)+f(0))^{2}$$