Question

Let $f(x)=x^6+2x^4+x^3+2x+3,x\in$ R. Then the natural number $n$ for which $\underset{x\to 1}{\lim }\frac{x^{n}f(1)-f(x)}{x-1}=44$ is _____

Answer 1

Answer: 7

$f(n)=x^6+2x^4+x^3+2x+3$
$\underset{x\to 1}{\lim }\frac{x^{n}f(1)-f(x)}{x-1}=44$
$\underset{x\to 1}{\lim }\frac{9x^n-\left(x^6+2x^4+x^3+2x+3\right)}{x-1}=44$
$\underset{x\to 1}{\lim }\frac{9n{x}^{n-1}-\left(6x^5+8x^3+3x^2+2\right)}{1}=44$
$\Rightarrow 9n-(19)=44$
$\Rightarrow 9n=63$
$\Rightarrow n=7$