Question

If a and b are positive numbers and $\underset{x\to 0}{lim}\frac{(1+a^3)+8e^{\frac{1}{x}}}{1+(2+b+b^2)e^{\frac{1}{x}}}=2$, then value of $a^2+b^2=$

Answer 1

$\underset{x\to 0}{lim}\frac{(1+a^3)+8e^{\frac{1}{x}}}{1+(2+b+b^2)e^{\frac{1}{x}}}=2$
$\underset{x\to 0}{lim}\frac{e^{-\frac{1}{x}}(1+a^3)+8}{e^{-\frac{1}{x}}+(2+b+b^2)}=2$
$\Rightarrow b^2+b-2=0$
$\Rightarrow b=-2,1$
$\Rightarrow b=1$ (since, given b is positive)
Also, since, limit of the function exists as its given as $2$
So, $LHL=2$
$\Rightarrow \underset{h\to 0}{lim}\frac{(1+a^3)+8e^{-\frac{1}{h}}}{1+(2+b+b^2)e^{-\frac{1}{h}}}=2$
$\Rightarrow \frac{(1+a^3)}{1}=2$
$\Rightarrow a=1$
Hence, $a^2+b^2=1+1=2$