Question

The value of $\underset{x\to 0}{\lim }\frac{1-\cos x+2\sin x-{\sin }^{3}x-x^2+3x^4}{{\tan }^{3}x-6{\sin }^{2}x+x-5x^3}$ is

(a) 1

(b) 2

(c) −1

(d) −2

Answer 1

(b) 2

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\left(1-\cos x\right)+2\sin x-{\sin }^{3}x-x^2+3x^4}{{\tan }^{3}x-6{\sin }^{2}x+x-5x^3} \\ \mathrm{Divi}\text{ding}{\mathrm{N}}^r\text{and}{\mathrm{D}}^r\mathrm{by}x: \\ \underset{x\to 0}{\lim }\frac{{\displaystyle \frac{2{\sin }^2{\displaystyle \frac{x}{2}}}{x}}+{\displaystyle \frac{2\sin x}{x}}-{\displaystyle \frac{{\sin }^{3}x}{x}}-x+3x^3}{{\displaystyle \frac{{\tan }^{3}x}{x}}-{\displaystyle \frac{6{\sin }^{2}x}{x}}+1-5x^2} \\ \underset{x\to 0}{\lim }\frac{{\displaystyle \frac{2x}{4}}{\displaystyle \frac{{\sin }^2{\displaystyle \frac{x}{2}}}{{\displaystyle \frac{x^2}{4}}}}+{\displaystyle \frac{2\sin x}{x}}-{\displaystyle \frac{x^2{\sin }^{3}x}{x^3}}-x+3x^3}{x^2\left({\displaystyle \frac{\tan 3x}{x^3}}\right)-{\displaystyle \frac{6x{\sin }^{2}x}{x^2}}+1-5x^2} \\ =\frac{0+2-0-0+0}{0-0+1-0}=2 \end{gathered}$$