Question

If $\underset{x\to 0}{lim}\left(\frac{cos4x+acos2x+b}{x^4}\right)$ is finite then the value of $a,b$ respectively are?

• A. $5,4$
• B. $-5,-4$
• C. $-4,3$
• D. $4,5$

Answer 1

Answer: (C)

$-4,3$

$\underset{x\to 0}{lim}\left(\frac{\cos 4x+a\cos 2x+b}{x^4}\right)$

$=\frac{1+a+b}{0^4}$

But it is given that value is finite

$\Rightarrow 1+a+b=0\to 1$

So that we get $\frac{0}{0}$form

$\underset{x\to 0}{lim}\frac{(-\sin 4x)\left(4\right)+a(-\sin 2x)\left(2\right)}{4x^3}$ (L-Hospital's Rule)

$\frac{0}{0}$form

$\underset{x\to 0}{lim}\frac{-\left(\cos 4x\right)\left(4\right)-a(+\cos 2x)}{3x^2}$

$\frac{-4-a}{3}$

but again the limit has finite value

$\Rightarrow -4-a=0$

$\Rightarrow a=-4$

$\Rightarrow b=4-1=3$ (from $1$)