Question

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

Answer 1

${lim}_{x\to 0}$ $(1+ax+b{x}^2{)}^{\frac{2}{x}}=e^3$ ....(given)

Taking logarithm both sides we get,

$log({lim}_{x\to 0}$ $(1+ax+b{x}^2{)}^{\frac{2}{x}})=log\left(e^3\right)$

Exchanging the limit and log functions we get,

${lim}_{x\to 0}$ $log(1+ax+b{x}^2{)}^{\frac{2}{x}}=3$

${lim}_{x\to 0}$ $\frac{2}{x}.log(1+ax+b{x}^2)=3$

${lim}_{x\to 0}$ $\frac{log(1+ax+b{x}^2)}{x}=\frac{3}{2}$

As left hind side limit is $\frac{0}{0}$ form so by applying L'Hospital's rule of limit we get,

${lim}_{x\to 0}$ $\frac{1}{(1+ax+b{x}^2)}$ $\times \frac{(a+2bx)}{1}=\frac{3}{2}$

Now by taking limit $x\to 0$ we get $a=\frac{3}{2}$

As final equation doesn't contain "b" that's why the limit doesn't delend on the value of "b" , so "b" can be any real number.

Hence $a=\frac{3}{2}$ and $bϵR$