Question

Let $f\left(x\right)=\frac{\tan x+a{e}^x+b{e}^{-x}+c\ln (1+x)}{x^3}$ and $a,b,c$ are real constants. If $\underset{x\to 0^{+}}{lim}f\left(x\right)$ is finite, then

• A. $a=-\frac{1}{2}$
• B. $b=\frac{1}{2}$
• C. $c=0$
• D. $\underset{x\to 0^{+}}{lim}f\left(x\right)=-\frac{1}{6}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $a=-\frac{1}{2}$
B $b=\frac{1}{2}$
C $c=0$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}\frac{\tan x+a{e}^x+b{e}^{-x}+c\ln (1+x)}{x^3}$

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

$=\underset{x\to 0^{+}}{lim}\frac{(a+b)+x(1+a-b+c)+x^2(\frac{a}{2}+\frac{b}{2}-\frac{c}{2})+x^3(\frac{1}{3}+\frac{a}{6}-\frac{b}{6}-\frac{c}{3})+x^4(...}{x^3}$
$=\underset{x\to 0^{+}}{lim}\frac{a+b}{x^3}+\frac{1+a-b+c}{x^2}+\frac{1}{x}(\frac{a}{2}+\frac{b}{2}-\frac{c}{2})+(\frac{1}{3}+\frac{a}{6}-\frac{b}{6}-\frac{c}{3})$

As $x\to 0$, denominator $\to 0$.
In order to limit exists, numerator $\to 0$
$\Rightarrow a+b=0\Rightarrow b=-a\Rightarrow 1+a-b+c=0\Rightarrow a+b-c=0\Rightarrow c=0$
$\Rightarrow a=-\frac{1}{2},b=\frac{1}{2}$

$\therefore \underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{1}{3}-\frac{1}{12}-\frac{1}{12}$
$=\frac{1}{6}$