How to prove that this function is increasing? $(a^x-1)/(x)$ is growing?, For $a>1, x>0$
I would like to know some way to prove that this function is increasing, without using derivation techniques, only with the basic calculation, using limits if necessary.
 A: If you are able to use the Maclaurin series for $\exp(x)$ then
$$a^x = \exp(x\ln a) = \sum_{n=0}^\infty \frac{(x\ln a)^n}{n!} = 1 + x\ln a + \frac{(x\ln a)^2}{2!} + \cdots$$
so
$$\frac{a^x-1}{x} = \ln a + \frac{(\ln a)^2}{2!}x + \frac{(\ln a)^3}{3!}x^2 + \cdots$$
Each non-constant term in this series is an increasing function of $x$ for $x>0$, so the entire function is too.
A: Strictly Increasing function for $f(x)$means $$\forall x_1,x_2$$ $$x_1 > x_2 => f(x_1) > f(x_2)$$
In your case, let $$f(x) = \frac{a^x - 1}{x} = t, f(x_1) = t_1, f(x_2) = t_2$$
Now,
$$a^x = xt + 1$$
$$=>x\ln{a} = \ln{(xt + 1)}$$
So,
$$x_1 > x_2$$
$$=>x_1\ln{a} > x_2\ln{a}$$
$$=>x_1\ln{a} > \ln(x_2t_2 + 1)$$
$$=>\ln(x_1t_1 + 1) > \ln(x_2t_2 + 1)$$
$$=>x_1t_1 + 1 > x_2t_2 + 1$$
$$=>x_1t_1 > x_2t_2$$.
$$=>(x_2+c)t_1 > x_2t_2$$, where $x_1 = x_2 + c, c > 0$
$$=>(x_2t_1 + ct_1) > x_2t_2$$
$$=>(x_2t_1 - x_2t_2) + ct_1) > 0$$
$$=>(x_2(t_1 - t_2) + ct_1) > 0$$
Now we know that $c > 0, t_1 = \frac{a^{x_1} - 1}{x_1} > 0$ because $a > 1, x >0$
so, $ct_1 > 0$
$$=>x_2(t_1 - t_2) > 0$$
Since $x_2 > 0 =>(t_1 - t_2) > 0 =>t_1 > t_2$,
So $f(x)$ is Strictly Increasing
