Question

If $\underset{x\to 3}{lim}{\left(\frac{\sqrt{2x+3}-x}{\sqrt{x+1}-x+1}\right)}^{\frac{x-1-\sqrt{x^2-5}}{x^2-5x+6}}$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ where $a,b,c\in N,$ then the least value of $a^2+b^2+c^2$ is

Answer 1

Let $L=\underset{x\to 3}{lim}\left(f\right(x{)}^{g\left(x\right)})$
$\Rightarrow \ln L=\underset{x\to 3}{lim}g\left(x\right)\cdot \ln \left(\underset{x\to 3}{lim}f\left(x\right)\right)$

$\underset{x\to 3}{lim}f\left(x\right)=\underset{x\to 3}{lim}\left(\frac{\sqrt{2x+3}-x}{\sqrt{x+1}-x+1}\right)$
Rationalising the numerator and denominator,
$\underset{x\to 3}{lim}f\left(x\right)$
$=\underset{x\to 3}{lim}\left(\frac{(\sqrt{2x+3}-x)(\sqrt{2x+3}+x)(\sqrt{x+1}+(x-1\left)\right)}{(\sqrt{x+1}-(x-1\left)\right)(\sqrt{2x+3}+x)(\sqrt{x+1}+(x-1\left)\right)}\right)=\underset{x\to 3}{lim}\left(\frac{(2x+3-x^2)(\sqrt{x+1}+(x-1\left)\right)}{(x+1-(x-1{)}^2\left)\right(\sqrt{2x+3}+x)}\right)=\underset{x\to 3}{lim}\frac{2}{3}\left(\frac{2x+3-x^2}{3x-x^2}\right)=\frac{2}{3}\underset{x\to 3}{lim}\left(\frac{x^2-2x-3}{x(x-3)}\right)=\frac{2}{3}\underset{x\to 3}{lim}\left(\frac{(x-3)(x+1)}{x(x-3)}\right)=\frac{2}{3}\left(\frac{4}{3}\right)=\frac{8}{9}$

Now, $\underset{x\to 3}{lim}g\left(x\right)$
$=\underset{x\to 3}{lim}\frac{x-1-\sqrt{x^2-5}}{x^2-5x+6}$
Again rationalising the numerator, we get
$\underset{x\to 3}{lim}g\left(x\right)$
$=\underset{x\to 3}{lim}\frac{(x-1-\sqrt{x^2-5})(x-1+\sqrt{x^2-5})}{(x^2-5x+6)(x-1+\sqrt{x^2-5})}=\underset{x\to 3}{lim}\frac{(x^2-2x+1-x^2+5)}{(x-2)(x-3)(x-1+\sqrt{x^2-5})}=\underset{x\to 3}{lim}\frac{-2(x-3)}{(x-2)(x-3)(x-1+\sqrt{x^2-5})}=\underset{x\to 3}{lim}\frac{-2}{(x-2)(x-1+\sqrt{x^2-5})}$
$=\frac{-2}{\left(1\right)\left(4\right)}=\frac{-1}{2}$

$\therefore L={\left(\frac{8}{9}\right)}^{-1/2}$
$=\frac{3}{2\sqrt{2}}=\frac{3\sqrt{2}}{4}$
Hence, the least value of $a^2+b^2+c^2$ is$3^2+2^2+4^2=29$