Question

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

• A. 44594
• B. 44563
• C. 44593
• D. 2, 0

Answer 1

Answer: (A)

44594

Answer 2

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