Question

$L{t}_{x\to 0}f\left(\frac{1-cos3x}{x^2}\right)$ where $f\left(x\right)$ is a continuous function satisfying the condition $f\left(\frac{9}{2}\right)=\frac{2}{9}$ is equal to:

• A. $0$
• B. $9/2$
• C. $2/9$
• D. $4$

Answer 1

Answer: (C)

$2/9$

$\cos x=\frac{1-x^2}{2!}+\frac{x^4}{4!}\therefore \cos \left(3x\right)=1-\frac{{\left(3x\right)}^2}{2!}+\frac{{\left(3x\right)}^4}{4!}={lim}_{x⟶0}f\left[\frac{1-(1-\frac{9x^2}{2!}+\frac{3^4{x}^4}{4!}....)}{x^2}\right]={lim}_{x⟶0}f\left[\frac{\frac{9x^2}{2!}+\frac{3^4{x}^4}{4!}....}{x^2}\right]={lim}_{x⟶0}f[\frac{9}{2!}+\frac{3^4{x}^4}{4!}+\frac{3^6{x}^4}{6!}...]=f[\frac{9}{2!}-0+\mathrm{0...}]=f\left[\frac{9x^2}{2!}\right]=f\left[\frac{9x^2}{2!}\right]=\frac{2}{9}$