Question

Let $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$. Then $\displaystyle\lim_{h \to0} \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,
$f(x)=3 x^{10}-7 x^{8}+5 x^{6}-21 x^{3}+3 x^{2}-7$
$\therefore f(1-h)=3(1-h)^{10}-7(1-h)^{8}$
$+5(1-h)^{6}-21(1-h)^{3}+3(1-h)^{2}-7$
$=3\left(1-10 h+45 h^{2}-120 h^{3}+\ldots \ldots+h^{10}\right)$
$-7\left(1-8 h+28 h^{2}-56 h^{3}+\ldots . .+h^{8}\right)$
$+5\left(1-6 h+15 h^{2}-20 h^{3}+\ldots . .+h^{6}\right)$
$ -21\left(1-3 h+3 h^{2}-h^{3}\right)$
$+3\left(-2 h+h^{2}\right)-7$
$\Rightarrow f(1-h)=-24+53 h+h^{2}(-46)+h^{3}(-47)+\ldots$
and $f(a)=-24$
$ \displaystyle\lim _{h \rightarrow 0} \frac{f(1-h)-f(x)}{h^{3}+3 h} $
$= \displaystyle\lim _{h \rightarrow 0} \frac{-24+53 h+h^{2}(-40)+h^{3}(-47)+\ldots-(-24)}{h\left(h^{2}+3\right)} $
$=\displaystyle\lim _{h \rightarrow 0} \frac{53 h+h^{2}(-46)+h^{3}(-47)+\ldots}{h\left(h^{2}+3\right)} $
$=\displaystyle\lim _{h \rightarrow 0} \frac{53+h(-40)+h^{2}(-47)+\ldots}{h^{2}+3}=\frac{53}{3}$