Question

$\underset{x\to 0}{lim}\frac{1-\cos (1-\cos x)}{x^4}$ is equal to

• A. $\frac{1}{8}$
• B. $-\frac{1}{8}$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

Answer: (A)

$\frac{1}{8}$

$A=\underset{x\to 0}{lim}\frac{1-\cos (1-\cos x)}{x^4}$

Let $1-\cos x=t$

$\Rightarrow$ when $x\to 0$, we also get $t\to 0$

$A=\underset{t\to 0}{lim}\frac{1-\cos t}{t^2}\times \underset{x\to 0}{lim}\frac{(1-\cos x{)}^2}{x^4}$

$=\underset{t\to 0}{lim}\frac{\sin t}{2t}\times \underset{x\to 0}{lim}{\left(\frac{1-\cos x}{x^2}\right)}^2$ ....... using L-hospital for first limit

Using $(1-\cos x)=2si{n}^2\frac{x}{2}$

$=1\times \underset{x\to 0}{lim}{\left(\frac{2si{n}^2\frac{x}{2}}{4{\left(\frac{x}{2}\right)}^2}\right)}^2$

$=\frac{1}{2}\times \frac{1}{2^2}$

$=\frac{1}{8}$