Question

If $\underset{x\to 0}{lim}\frac{a{e}^x-bcosx+c{e}^{-x}}{xsinx}=2$,

then $a+b+c=?$

• A. $2$
• B. $4$
• C. $0$
• D. $6$

Answer 1

The correct option is B $4$

We observe that as x $\to$ 0, the denominator tends to 0 and the numerator tends to $a-b+c$. Therefore, for the limit to exist we must have
$a-b+c=0$ ........ (i)
$\underset{x\to 0}{lim}\frac{a{e}^x-bcosx+c{e}^{-x}}{xsinx}$ is in the $\frac{0}{0}$ form.
$\therefore \underset{x\to 0}{lim}\frac{a{e}^x+bsinx-c{e}^{-x}}{sinx+xcosx}=2$ [Using L' Hospital's Rule]
Here, the numerator is tending to $a-c$ as $x\to$ 0. Therefore, for the limit to exist, we must have
$a-c=0$ ....... (ii)
$\underset{x\to 0}{lim}\frac{a{e}^x+bsinx-c{e}^{-x}}{sinx+xcosx}$ is in the form $\frac{0}{0}$
$\therefore \underset{x\to 0}{lim}\frac{a{e}^x+bcosx+c{e}^{-x}}{2cosx-xsinx}=2$
$\Rightarrow \frac{a+b+c}{2}=2$
$\Rightarrow a+b+c=4$