Question

Let $f(x)=3x^{10}-7x^8+5x^6-21x^3+3x^2-7$. Then $\underset{h\to 0}{\lim }\frac{f\left(1-h\right)-f\left(1\right)}{h^3+3h}$

• A. does not exsit
• B. is $\frac{50}{3}$
• C. is $\frac{53}{3}$
• D. is $\frac{22}{3}$

Answer 1

Answer: (C)

We have,
$f(x)=3x^{10}-7x^8+5x^6-21x^3+3x^2-7$
$\therefore f(1-h)=3(1-h{)}^{10}-7(1-h{)}^8$
$+5(1-h{)}^6-21(1-h{)}^3+3(1-h{)}^2-7$
$=3\left(1-10h+45h^2-120h^3+\dots \dots +h^{10}\right)$
$-7\left(1-8h+28h^2-56h^3+\dots ..+h^8\right)$
$+5\left(1-6h+15h^2-20h^3+\dots ..+h^6\right)$
$-21\left(1-3h+3h^2-h^3\right)$
$+3\left(-2h+h^2\right)-7$
$\Rightarrow f(1-h)=-24+53h+h^2(-46)+h^3(-47)+\dots$
and $f(a)=-24$
$\underset{h\to 0}{\lim }\frac{f(1-h)-f(x)}{h^3+3h}$
$=\underset{h\to 0}{\lim }\frac{-24+53h+h^2(-40)+h^3(-47)+\dots -(-24)}{h\left(h^2+3\right)}$
$=\underset{h\to 0}{\lim }\frac{53h+h^2(-46)+h^3(-47)+\dots }{h\left(h^2+3\right)}$
$=\underset{h\to 0}{\lim }\frac{53+h(-40)+h^2(-47)+\dots }{h^2+3}=\frac{53}{3}$