Question

Let $y=f(x)$ be an infinitely differentiable function on $R$ such that $f (0) \neq 0$ and $\frac{ d ^{ n } y }{ dx ^{ n }} \neq 0$ at $x =0$ for $n =1,2,3,4$. If $\underset{x \rightarrow 0}{\text{Lim}} \frac{f(4 x)+a f(3 x)+b f(2 x)+c f(x)+d f(0)}{x^4}$ exists, then find the value of $(25 a+50 b+100 c+500 d)$.

Answer 1

The fact that the limit exists implies that
$\underset{x \rightarrow 0}{\text{Lim}}(f(4 x)+a f(3 x)+b f(2 x)+c f(x)+d f(0))=(1+a+b+c+d) f(0)=0$
$\therefore a + b + c + d =-1$ ....(1)
Apply L'Hospital Rule once, then we have
$\underset{x \rightarrow 0}{\text{Lim}}\frac{f(4 x)+a f(3 x)+b f(2 x)+c f(x)+d f(0)}{x^4}=\underset{x \rightarrow 0}{\text{Lim}}\frac{4 f^{\prime}(4 x)+3 a f^{\prime}(3 x)+2 b f^{\prime}(2 x)+c f^{\prime}(x)}{4 x^3}$
and for the following limit to exist, we also need
$\underset{x \rightarrow 0}{\text{Lim}}\left(4 f^{\prime}(4 x)+3 a f^{\prime}(3 x)+2 b f^{\prime}(2 x)+c f^{\prime}(x)\right)=(4+3 a+2 b+c) f^{\prime}(0)=0, $
\therefore & 3 a+2 b+c=-4 \ldots .(2)
Repeat this process twice and get another two equations as
$9 a +4 b + c =-16$....(3)
and $27 a+8 b+c=-64$ ....(4)
Now, (4) - (3) $\Rightarrow 18 a +4 b +48=0 \Rightarrow 9 a +2 b +24=0$...(5)
(3) $-(2) \Rightarrow 6 a +2 b +12=0 \Rightarrow 6 a +2 b +12=0$....(6)
(5) $-6(6) \Rightarrow 3 a+12=0 \Rightarrow a=-4, b=6$
From equation (2), we get $c=-4$
and from equation (1), $d =1$.
Hence $(25 a +50 b +100 c +500 d )=-100+300-400+500=300$