Question

$\underset{x\to 0}{lim}\frac{a\cos x+bx\sin x-5}{x^4}=$, finite, find $a$ and $b$ and the limit.

• A. $a=5$
• B. $b=\frac{5}{2}$
• C. limit$=\frac{5}{24}$
• D. $a=\frac{5}{2}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a=5$
B $b=\frac{5}{2}$
C limit$=\frac{5}{24}$

$L=\underset{x\to 0}{lim}\frac{a\cos x+bx\sin x-5}{x^4}=\underset{x\to 0}{lim}\frac{1}{x^4}\{a[1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...\infty ]+bx[x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\infty ]-5\}$

$L=\underset{x\to 0}{lim}\{a[\frac{1}{x^4}-\frac{1}{x^{2}2!}+\frac{1}{4!}]+b[\frac{1}{x^2}-\frac{1}{3!}]-\frac{5}{x^4}\}=\underset{x\to 0}{lim}\{\frac{(a-5)}{x^4}+\frac{(b-2a)}{{2x}^2}+\frac{a}{4!}-\frac{b}{3!}\}$

L is finite when $a-5=0$ and $2a-b=0$
Therefore, $a=5$ and $b=5/2$
now, $L=\underset{x\to 0}{lim}\{\frac{a}{4!}-\frac{b}{3!}\}=\frac{5}{4!}-\frac{5}{2.3!}=\frac{5}{24}$
Ans: A,B,C