Question

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

Answer 1

Answer: 300

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