Ever notice that the graphs of [math]\sqrt{x}[/math] and [math]\ln x[/math] are similar and wonder why? Adjust the sliders for [math]a[/math], [math]b[/math], and [math]n[/math] in the app below in small increments to see the effects on the graph of [math]F\left(x\right)=ax^{{1}/{n}}-b[/math]. The initial values are [math]a=1[/math], [math]b=0[/math], and [math]n=2[/math], so [math]F\left(x\right)=\sqrt{x}[/math] After a few observations, check the [i]set a=n [/i]box so that [math]a[/math], [math]b[/math] and [math]n[/math] move together and [math]F\left(x\right)=nx^{1/n}-n=n\left(x^{1/n}-1\right)[/math]. Observe what happens to the graph of [math]F\left(x\right)[/math] in relation to the graph of [math]\ln x[/math] with each change.

Recall that [math]\int x^rdx=\frac{1}{r+1}x^{r+1}+C[/math] for any rational number [math]r[/math], except [math]r=-1[/math], where [math]\int x^{-1} \ dx=\ln|x|+C.[/math] [br][br]Here we are only interested in positive values of [math]x,[/math] so [math]\int x^{-1} = \ln x +C, \ \ x>0.[/math][br][br]Next, since [math]f(x)=x^{-1+1/n}[/math] is continuous for [math]x>0,[/math] we can assume that [br][br][center][math]\lim_{n \to \infty} \int x^{-1+1/n} \ dx +C_1 = \int \lim_{n \to \infty} x^{-1+1/n} \ dx[/math]. (1)[/center](Note: this assumption works here, but is not necessarily true in general.)[br][br]The right side of equation (1) can be rewritten as [math]\int \lim_{n \to \infty} x^{-1+1/n} \ dx =\int x^{-1} \ dx = \ln x+C_2.[/math]. [br][br]Substituting for the right side of equation (1) and combining constants, [br][br][center][math]\begin{align}\lim_{n \to \infty} \int x^{-1+1/n} \ dx +C_3 & = \ln x \\[br] \lim_{n \to \infty} \left( nx^{1/n} + C_4 \right)+ C_3 &= \ln x\\[br]\lim_{n \to \infty} \left( nx^{1/n} +C_4+ C_3\right) &= \ln x\\[br]\lim_{n \to \infty} \left( nx^{1/n} +C \right)&= \ln x[br]\end{align}[/math] [br][/center][br]Because [math]\ln1=0[/math], choose the constant of integration, [math]C,[/math] so that [math]nx^{1/n}+C =0[/math] when [math]x=1.[/math][br][br][math]0 = n(1)^{1/n}+C \ \ \ \Longrightarrow \ \ \ C = -n[/math][br][br]Choosing [math]C=-n[/math], [br][br][math]\huge \ln x = \lim_{n \to \infty } (nx^{1/n}-n) [/math]