Question

If $\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}=2$, then

• A. $a=c$
• B. $c=1$
• C. $a=2$
• D. $b=2$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $a=c$
B $c=1$
D $b=2$

We have $\underset{x\to 0}{lim}\frac{a{e}^x-b\cos x+c{e}^{-x}}{x\sin x}$

$=\underset{x\to 0}{lim}\frac{a(1+x+\frac{x^2}{2!}+...)-b(1-\frac{x^2}{2!}+...)+c(1-x+\frac{x^2}{2!}+...)}{x(x-\frac{x^3}{3!}+...)}$

$=\underset{x\to 0}{lim}\frac{(a-b+c)+(a-c)x+x^2(\frac{a}{2!}+\frac{b}{2!}+\frac{c}{2!})+0\left(x^3\right)}{x^2(1-\frac{x^2}{3!}+\frac{x^4}{5!}-...)}$
This limit exists if $a-b+c=0,a-c=0$ and is equal to 2 and if $(a+b+c)/2=2,$ i.e. $a+b+c=4$.

Solving these equations, we get $a=1,c=1$ and $b=2$.
We can apply L-Hospital's rule also to get the required conclusion.