Question

If $a=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{1}{(r+2)r!}$ and $b=\underset{x\to 0}{lim}\frac{e^{\sin x}-e^x}{\sin x-x}$, then `

• A. $a=b$
• B. $a=2b$
• C. $2a=b$
• D. $a+b=0$

Answer 1

Answer: (C)

$2a=b$

Let $t_r=\frac{1}{(r+2)r!}=\frac{r+1}{(r+2)!}+\frac{(r+2)-1}{(r+2)!}$

$\Rightarrow t_r=\frac{1}{(r+1)!}-\frac{1}{(r+2)!}$

Now, $\sum _{r=1}^n{t}_r=(\frac{1}{2!}-\frac{1}{3!})+(\frac{1}{3!}-\frac{1}{4!})t_r=+(\frac{1}{4!}-\frac{1}{5!})+...+(\frac{1}{(n+1)!}-\frac{1}{(n+2)!})$

$\Rightarrow \sum _{r=1}^n{t}_r=\frac{1}{2!}-\frac{1}{(n+2)!}$

Hence, $a=\underset{n\to \infty }{lim}[\frac{1}{2}-\frac{1}{(n+2)!}]=\frac{1}{2}$

Now, $b=\underset{x\to 0}{lim}\frac{e^{\sin x}-e^x}{\sin x-x}$

$\Rightarrow b=\underset{x\to 0}{lim}{e}^x.\left(\frac{e^{\sin x-x}-1}{\sin x-x}\right)=e^0.1=1$

$\therefore b=2a$