Question

The value of ${lim}_{x\to 0}\frac{(1-cosx)+2sinx-si{n}^{3}x-x^2+3x^4}{ta{n}^{3}x-6si{n}^{2}x+x-5x^3}$ is

• A. 1
• B. 2
• C. -1
• D. -2

Answer 1

The correct option is B

2

${lim}_{x\to 0}\frac{(1-cosx)+2sinx-si{n}^{3}x-x^2+3x^4}{ta{n}^{3}x-6si{n}^{2}x+x-5x^3}$
Dividing N' and D' by x :
$={lim}_{x\to 0}\frac{2si{n}^2\frac{\frac{x}{2}}{x}+\frac{2sinx}{x}-\frac{si{n}^{2}x}{x}-x+3x^3}{\frac{ta{n}^{3}x}{x}-\frac{6si{n}^2}{x}+1-5x^2}$
$={lim}_{x\to 0}\frac{\frac{2x}{4}\frac{si{n}^2\frac{x}{2}}{\frac{x^2}{4}}+\frac{2sinx}{x}-\frac{x^{2}si{n}^{3}x}{x^3}-x+3x^3}{x^3\left(\frac{tan3x}{x^3}\right)-\frac{6si{n}^{2}x}{x^2}+1-5x^2}$
$=\frac{0+2-0-0+0}{0-0+1-0}=2$