Question

If ${lim}_{x\to 1}(1+ax+b{x}^2{)}^{\frac{c}{x-1}}=e^3$, then find conditions on a, b and c.

Answer 1

Consider the given function

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

We know that,

$e^{\log x}=x$

Then,

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

Comparing both side and we get,

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

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

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

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

Applying L’ Hospital rule and we get,

${lim}_{x\to 1}\frac{c\left(\frac{1}{1+ax+b{x}^2}\right)(0+a+2bx)}{1-0}=0$

Taking limit and we get,

$c\left(\frac{1}{1+a\times 1+b{\left(1\right)}^2}\right)(0+a+2b\times 1)=0$

$ac+2bc=0$

Hence, this is the answer.