Question

If $f\left(x\right) = 3x^{10} - 7x^{8} + 5x^{6} -21x^{3} + 3x^{2}-7$, then $ \displaystyle\lim_{\alpha\to0} \frac{f\left(1- \alpha\right) - f\left(1\right)}{\alpha^{3} + 3 \alpha} $ is

• A. $ - \frac{53}{3}$
• B. $ \frac{53}{3}$
• C. $ - \frac{55}{3}$
• D. $ \frac{55}{3}$

Answer 1

Answer: (B)

Let $f\left(x\right) = 3x^{10} - 7x^{8} + 5x^{6} -21x^{3} + 3x^{2}-7$
$ f'\left(x\right)=30x^{9} - 56x^{7}+30x^{5}-63x^{2}+6x$
$ f' \left(1\right) = 30 -56 +30-63+6$
$ = 66 -63 - 56 =-53 $
Consider $\displaystyle\lim_{\alpha \to 0} \frac{f (1 - \alpha) - f(1)}{\alpha^3 + 3\alpha}$
$ = \displaystyle\lim_{\alpha\to0} \frac{f'\left(1-\alpha\right)\left(-1\right)-0}{3\alpha^{2} +3}$ (By using L’hospital rule)
$ =\frac{ f'\left(1-0\right)\left(-1\right)}{3\left(0\right)^{2} + 3} = \frac{-f'\left(1\right)}{3}= \frac{53}{3}$