Question

Evaluate ${lim}_{x\to 3\pi /4}\frac{1+\sqrt[3]{3\tan x}}{1-2{\cos }^{2}x}$

• A. 1/4
• B. -1/4
• C. 1/2
• D. -1/2

Answer 1

The correct option is B -1/4

${lim}_{x\to 3\pi /4}\frac{1+\sqrt[3]{3\tan x}}{1-2{\cos }^{2}x}$

it is the form $\frac{0}{0}$

So, we will apply L Hospital Rule,

${lim}_{x\to 3\pi /4}\frac{\frac{d}{dx}(1+\sqrt[3]{3\tan x})}{\frac{d}{dx}(1-2{\cos }^{2}x)}$

${lim}_{x\to 3\pi /4}\frac{tan{x}^{-2/3}.se{c}^{2}x}{-4cosx(-sinx)}={lim}_{x\to 3\pi /4}\frac{tan(3\pi /4{)}^{-2/3}.se{c}^2(3\pi /4)}{-4cos\left(3\pi /4\right)(-sin(3\pi /4\left)\right)}=\frac{(-1{)}^{-2/3}.\frac{1}{2}}{-4\frac{1}{2}}=-\frac{1}{4}$