Square root of 2=1.41421..... Cube root of 3=1.44224..... 4th root of 4=1.41421...... 5th root of 5=1.37972...... 6th root of 6=1.34800...... 7th root of 7=1.32046...... . . nth root of n=.......... Similarly thereafter nth root of n will converge towards 1.
Q.2 why is square root 2 equals 4th root of 4? Intuitively sqrt 2 is diagnol of 1 by 1 square, and sqrt 3 is diagnol of 1 by 1 parallelogram, explain geometrially?
Let us find extreme values of $f(x)=x^{\frac1x}$ where $x$ is real positive
So, $\ln f(x)=\frac{\ln x}x$
$$\frac{d \ln f(x)}{dx}=\frac{1-\ln x}{x^2}$$
So, for the extreme values of $\ln f(x),1-\ln x=0\iff x=e$
$$\frac{d^2 \ln f(x)}{dx^2}=\frac{-\frac1x\cdot x^2-(1-\ln x)(2x)}{x^4}$$ which is $<0$ at $x=e$
and $$\frac{d \ln f(x)}{dx}>0\text{ or } < \text{ according as }1-\ln x>0\iff x<e\text{ or }1-\ln x<0\iff x>e $$
So, $x^{\frac1x}$ will increase when $x\in(0,e)$ and decreasing when $x\in(e,\infty)$ and we know $2<e<3$
As $e$ lies between $2,3$ the above derivation can not help us to compare $x^{\frac1x}$ at $x=2,3$ or any two values between which $e$ lies
So we can solve it as follows:
$\sqrt2 <\sqrt[3]3 $ i.e., $2^{\frac12}<3^{\frac13}\iff (2^{\frac12})^6<(3^{\frac13})^6\iff 2^3<3^2$ which is true
We have taken $6$th power as lcm$(2,3)=6$