NIELIT 2016 MAR Scientist C - Section B: 13

Question

$\displaystyle \lim_{x \rightarrow 0}\frac{1}{x^{6}} \displaystyle \int_{0}^{x^{2}}\frac{t^{2}dt}{t^{6}+1}=$?

• A. $1/4$
• B. $1/3$
• C. $1/2$
• D. $1$

Answer 1

take t^3 =y then solve u wil get ans as option B 

Answer 2

Option B is correct

Substitute  $t=\left ( tan\theta \right )^{\frac{1}{3}}$  ------(1)

$\Rightarrow$  $dt=\frac{1}{3}\times(tan \theta)^{\frac{-2}{3}}\times sec^2\theta$  $d\theta$

$\therefore$ The integral would now become   $\frac{(tan \theta)^{\frac {2}{3}}\times \frac{1}{3}\times (tan \theta)^{\frac{-2}{3}}\times sec^2\theta}{ 1+tan^2\theta }$   $d\theta$

$\Rightarrow$   $\frac{1}{3}\times$$\int_{0}^{x^2} \theta$  $d\theta= \frac{1}{3}\times$  $\frac{tan^{-1}(x^{6})}{x^{6}}$     (using identity $1+tan^{2}\theta=sec^{2}\theta$ and resubstituting value of $t$ from equation 1)

$\therefore$ limit now becomes  $\lim_{x \to0}$   $\frac{1}{3}\times \frac{tan^{-1}(x^{6})}{x^{6}}$

 if  $x\rightarrow0$  then  $x^{6}\rightarrow 0$ .  So, using the formula  $\lim_{x \to0}$  $\frac{tan^{-1}(x)}{x}=1$, we get the answer as  $\frac{1}{3}\times 1=\frac{1}{3}$

Comments

Modified and used $\tan \theta.$