Ludicrous Limit

The limit is of an indeterminate form, i.e., $1^{ \infty}$ . We can rewrite the limit as $e^P$ where $P= \lim_{x \to \infty} (\frac{a^{\frac{1}{x}} + b^{\frac{1}{x}}}{2} - 1)x$ .

Put $x= \frac{1}{t}$ to get $P= \lim_{t \to 0} (\frac{a^t +b^t -2}{2t})$ which is $\frac{0}{0}$ form.

Applying L'Hospital 's rule, we get: $P= \lim_{t \to 0} (\frac{a^t ln(a)+b^t ln(b) }{2}) = ln(\sqrt{ab})$

So, $e^P = e^{ln(\sqrt{ab})} = \sqrt{ab} = \sqrt{3 \times 12} = 6$ .

$\Large\begin{array}{rcl} f(12,3)&=& e^{\lim\limits_{x\rightarrow\infty}x[\ln(12^{\frac1x}+3^{\frac1x})-\ln2]}\\ &=& e^{\lim\limits_{h\rightarrow0}\frac{\ln(12^h+3^h)-\ln2}h}\\ &=& e^{\lim\limits_{h\rightarrow0}\frac{\ln3^h+\ln(4^h+1)-\ln2}h}\\ &=& e^{\lim\limits_{h\rightarrow0}\frac{\ln3^h}h+\frac{\ln(4^h+1)-\ln2}h}\\ &=& e^{\ln3+\lim\limits_{h\rightarrow0}\frac{(\ln4)4^h}{4^h+1}}\\ &=& e^{\ln3+\ln4-\lim\limits_{h\rightarrow0}\frac{\ln4}{4^h+1}}\\ &=& e^{\ln3+\ln4-\frac{\ln4}2}\\ &=& e^{\ln3+\ln4-\ln2}\\ &=& e^{\ln6}\\ &=&6 \end{array}$

Use AM>=G.M

Using Generalised Means we note that the limit can be written as $\lim_{x\to0}\left(\frac{a^x+b^x}{2}\right)^{1/x}$ Which we know is the Geometric Mean of two numbers. Therefore taking the geometric mean of $12$ and $3$ we have $\sqrt{12\times3}=\sqrt{36}=\boxed{6}$ .

$\lim _{ x\rightarrow \infty }{(\frac{a^\frac{1}{x} + b^\frac{1}{x}}{2})^x }$

$= \lim _{ x\rightarrow \infty }{(1+\frac{a^\frac{1}{x} + b^\frac{1}{x}-2}{2})^x }$

$= e^l$ where $l = \lim _{ x\rightarrow \infty }{ \frac { ({ a }^{ \frac { 1 }{ x } }+{ b }^{ \frac { 1 }{ x } }-2)x }{ 2 } }$

Put $y = \frac{1}{x}$

$l = \lim _{ y\rightarrow 0 }{ \frac { ({ a }^{ y }+{ b }^{ y }-2) }{ 2y } }$

$= \frac{1}{2}[ \lim _{ y\rightarrow 0 }{\frac{a^y - 1}{y}} + \lim _{ y\rightarrow 0 }{\frac{b^y - 1}{y}}]$

= $\frac{1}{2}\ln { ab }$

$\Rightarrow e^l = \sqrt { ab } = \sqrt{12*3} = 6$