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If $\lim _{x \rightarrow 0} \frac{a x e^{x}-b \log (1+x)}{x^{2}}=3$ then the values of $a$ and $b$ are, respectively
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The correct answer is:
2,2
We have. $\lim _{x \rightarrow 0} \frac{\operatorname{ax} e^{x}-b \log (1+x)}{x^{2}}=3\left[\frac{0}{0}\right.$ form $]$
Using L' Hospital's rule, we get $\lim _{x \rightarrow 0} \frac{a e^{x}+a x e^{x}-\frac{b}{1+x}}{2 x}=3 \quad\left[\frac{0}{0}\right.$ form $]$
$\Rightarrow a-b=0 \Rightarrow a=b$
Using L' Hospital's rule, we get
$$
\lim _{x \rightarrow 0} \frac{a e^{x}+a e^{x}+a x e^{x}+\frac{b}{(1+x)^{2}}}{2}=3
$$
$\Rightarrow \quad \lim _{x \rightarrow 0} 2 a e^{x}+a x e^{x}+\frac{b}{(1+x)^{2}}=6$
$\Rightarrow$
$2 a+b=6$
$\Rightarrow \quad 3 a=6 \Rightarrow a=2$
On putting the value of $a$ we get $b=2$
Using L' Hospital's rule, we get $\lim _{x \rightarrow 0} \frac{a e^{x}+a x e^{x}-\frac{b}{1+x}}{2 x}=3 \quad\left[\frac{0}{0}\right.$ form $]$
$\Rightarrow a-b=0 \Rightarrow a=b$
Using L' Hospital's rule, we get
$$
\lim _{x \rightarrow 0} \frac{a e^{x}+a e^{x}+a x e^{x}+\frac{b}{(1+x)^{2}}}{2}=3
$$
$\Rightarrow \quad \lim _{x \rightarrow 0} 2 a e^{x}+a x e^{x}+\frac{b}{(1+x)^{2}}=6$
$\Rightarrow$
$2 a+b=6$
$\Rightarrow \quad 3 a=6 \Rightarrow a=2$
On putting the value of $a$ we get $b=2$
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