Mathematics - Limits Question with Solution | TestHub

Let $f\left(x\right)=\left(x-1\right)\left(x^4+a{x}^3+b{x}^2+cx+d\right)$

$f\left(x\right)+f^{\text{'}}\left(x\right)+f^{\text{'}\text{'}}\left(x\right)=x^5+\left(a+4\right)x^4+\left(b+3a+16\right)x^3+$

$\left(9a+2b+c-12\right)x^2+\left(d+c+4b-6a\right)x+\left(c-2b\right)$

$\because f\left(x\right)+f^{\text{'}}\left(x\right)+f^{\text{'}\text{'}}\left(x\right)=x^5+64$

$\Rightarrow a+4=0, b+3a+16=0, 9a+2b+c-12=0, d+c+4b-6a=0$

$\Rightarrow a=-4, b=-4, c=56, d=-64$

So $\underset{x\to 1}{\lim }\frac{f\left(x\right)}{x-1}=\underset{x\to 1}{\lim }\left(x^4-4x^3-4x^2+56x-64\right)$

$=1-4-4+56-64=-15$