Discussion Forum

1)

Let  $\alpha,\beta\epsilon R$   be such that $\lim_{x \rightarrow0}\frac{x^{2}\sin (\beta x)}{\alpha x-\sin x}=1$   Then, 

 $6(\alpha+\beta)$  equals 

Answer:

Option D

Explanation:

Here,   $\lim_{x \rightarrow0}\frac{x^{2}\sin (\beta x)}{\alpha x-\sin x}=1$

$\Rightarrow$    $\lim_{x \rightarrow 0}\frac{x^{2}\left[ \beta x- \frac{(\beta x)^{3}}{3!}+\frac{(\beta x)^{5}}{5!}-.....\right]}{\alpha x-\left[ x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-......\right]}$ =1

  $\Rightarrow$   $\lim_{x \rightarrow 0}\frac{x^{3}\left[\beta-\frac{\beta^{3}x^{2}}{3!}+\frac{\beta^{5}x^{4}}{5!}-.....\right]}{(\alpha-1)x+\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-....}=1$

Limits exist only,

   when  , $\alpha -1=0$

 $\Rightarrow$    $\alpha =1$             ............(i)

$\therefore$    $\lim_{x \rightarrow 0}\frac{x^{3}\left[ \beta-\frac{\beta^{3}x^{2}}{3!}+\frac{\beta^{5}x^{4}}{5!}- .....\right]}{x^{3\left[\frac{1}{3!}-\frac{x^{2}}{5!}-  ....\right]}}=1$

 $\Rightarrow$    $6\beta=1$          ........(ii)

  From Eq .(i) and (ii),  we get   $6(\alpha+\beta)=6\alpha+6\beta$

                      =6+1= 7