Question

$li{m}_{x\to 0}\frac{(1-cos2x)(3+cosx)}{xtan4x}$ is equal to

• A. 4
• B. 3
• C. 2
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

We have $li{m}_{x\to 0}\frac{(1-cos2x)(3+cosx)}{xtan4x}=li{m}_{x\to 0}\frac{2si{n}^{2}x(3+cosx)}{x\times \frac{tan4x}{4x}\times 4x}$
= $li{m}_{x\to 0}\frac{2si{n}^{2}x}{x^2}\times li{m}_{x\to 0}\frac{(3+cosx)}{4}\times \frac{1}{li{m}_{x\to 0}\frac{tan4x}{4x}}$
= $2\times \frac{4}{4}\times 1$ $[\because li{m}_{heta\to 0}\frac{sinheta}{heta}=1andli{m}_{heta\to 0}\frac{tan}{heta}=1]$
= 2.