Question

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

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

Answer 1

Answer: (A)

We have,
$\underset{x\to 0}{\lim }\frac{{\text{axe}}^x-b\log (1+x)}{x^2}=3[\frac{0}{0}$ form $]$
Using L' Hospital's rule, we get
$\underset{x\to 0}{\lim }\frac{a{e}^x+{\text{axe}}^x-\frac{b}{1+x}}{2x}=3[\frac{0}{0}$ form $]$
$\Rightarrow a-b=0\Rightarrow a=b$
Using L' Hospital's rule, we get
$\underset{x\to 0}{\lim }\frac{a{e}^x+a{e}^x+{\text{axe}}^x+\frac{b}{(1+x{)}^2}}{2}=3$
$\Rightarrow \underset{x\to 0}{\lim }2a{e}^x+ax{e}^x+\frac{b}{(1+x{)}^2}=6$
$\Rightarrow 2a+b=6$
$\Rightarrow 3a=6$ ...(i)
$\Rightarrow a=2$
On putting the value of $a$ in Eq. (i),we get $b=2$