Question

If $\underset{x\to 0}{lim}\frac{ax{e}^x-b\log (1+x)}{x^2}=3$ then the values of $a$ and $b$ are, respectively

• A. $2,2$
• B. $1,2$
• C. $2,1$
• D. $2,0$

Answer 1

The correct option is A $2,2$

We have,
$\underset{x\to 0}{lim}\frac{ax{e}^x-b\log (1+x)}{x^2}=3\left[\frac{0}{0}\text{form}\right]$

Using 'L' Hospital's rile, we get
$\underset{x\to 0}{lim}\frac{a{e}^x+ax{e}^x-\frac{b}{1+x}}{2x}=3\left[\frac{0}{0}\text{form}\right]$
$\Rightarrow a-b=0\Rightarrow a=b$

Using L'Hospital's rule, we get
$\underset{x\to 0}{lim}\frac{a{e}^x+a{e}^x+ax{e}^x+\frac{b}{(1+x{)}^2}}{2}=3$

$\Rightarrow \underset{x\to 0}{lim}2a{e}^x+ax{e}^x+\frac{b}{(1+x{)}^2}=6$

$\Rightarrow 2a+b=6$
$\Rightarrow 3a=6\Rightarrow a=2$

On putting the value of a Eq. (i), we get $b=2$