Question

The value of the limits, $\underset{x\to 0}{lim}\frac{(sinx-x{)}^2+1-cos{x}^3}{x^{5}sinx+(1-cosx{)}^2(2x^2-sin{x}^2)}$ is equal to

• A. $\frac{19}{45}$
• B. $\frac{4}{9}$
• C. $\frac{19}{25}$
• D. $0$

Answer 1

Answer: (B)

$\frac{4}{9}$

$\underset{x\to 0}{lim}\frac{(sinx-x{)}^2+1-cos{x}^3}{x^{5}sinx+(1-cosx{)}^2(2x^2-sin{x}^2)}$
$=\underset{x\to 0}{lim}\frac{x^2(\frac{sinx}{x}-1{)}^2+2{\sin }^2\frac{x^3}{2}}{x^6\frac{sinx}{x}+\left(2{\sin }^2\frac{x}{2}{)}^2{x}^2\right(2-\frac{sin{x}^2}{x^2})}$ $\dots [\because cos2\theta =1-2si{n}^2\theta ]$
$=\underset{x\to 0}{lim}\frac{x^2(\frac{sinx}{x}-1{)}^2+2\frac{{\sin }^2\frac{x^3}{2}}{\frac{x^6}{4}}\frac{x^6}{4}}{x^6\frac{sinx}{x}+2\frac{x^4}{16}\frac{{\sin }^4\frac{x}{2}}{\frac{x^4}{16}}{x}^2(2-\frac{sin{x}^2}{x^2})}$
$=\underset{x\to 0}{lim}\frac{0+\frac{x^6}{2}}{x^6+\frac{x^6}{8}}$

$=\underset{x\to 0}{lim}\frac{\frac{x^6}{2}}{\frac{9x^6}{8}}$

$=\underset{x\to 0}{lim}\frac{x^6}{2}\times \frac{8}{9x^6}$

$=\frac{8}{2.9}$

Hence, option 'B' is correct.

$=\frac{4}{9}$