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$(a^x-1)/(x)$ 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.

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    $\begingroup$ In shortcut,take an arbitrary value of $a$.say,$a=5$,now,put x1=2 and calculate y1,then put x2=3 and calculate y2,as we are seeing clearly x2>x1,so if y2>y1,then the function is increasing.... $\endgroup$ – Rakibul Islam Prince Sep 14 '18 at 17:21
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    $\begingroup$ @RakibulIslamPrince How is a single example a valid "shortcut" to a proof? $\endgroup$ – John Coleman Sep 14 '18 at 19:46
  • $\begingroup$ I haven't titled it as a proof ,sir.I have just shown it as a way to see the function is increasing. $\endgroup$ – Rakibul Islam Prince Sep 14 '18 at 19:52
  • $\begingroup$ @RakibulIslamPrince But that is not a way to see that the function is increasing. At best, you see something which is consistent with the claim being true, but which is also consistent with the claim being false. $\endgroup$ – John Coleman Sep 14 '18 at 19:58
  • $\begingroup$ Sir, it is basically working as a power function,as $x$ is increasing and $x>0$ the function gets higher and higher more.It is also visible from it's graph. $\endgroup$ – Rakibul Islam Prince Sep 14 '18 at 20:04
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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

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  • $\begingroup$ I'm not sure about the last inequality $t_1>t_2$. Counter-example: $x_1=3, t_1=4, x_2=2, t_2=5$. Then $x_1 t_1=3\times4>2\times5=x_2 t_2$ and $x_1=3>2=x_2$ but it does not follow that $t_1=4>5=t_2$. $\endgroup$ – Matt Sep 14 '18 at 21:16
  • $\begingroup$ $t_1$ in this case cannot be 4 since it is $\frac{a^3 -1}{3}$. it depends on $a$. Note that $t_1 = f(x_1)$ $\endgroup$ – Gopal Anantharaman Sep 14 '18 at 22:11
  • $\begingroup$ The point is that $t_1>t_2$ does not follow from $x_1 t_1 > x_2 t_2 \wedge x_1>x_2$. Your argument assumes this to always be the case. $\endgroup$ – Matt Sep 14 '18 at 22:19
  • $\begingroup$ $x_2(t_1-t_2)>0$ still does not follow. $\endgroup$ – Matt Sep 15 '18 at 14:04
  • $\begingroup$ Thanks! Yes. I see your point. Let me see if I can find an easier proof. $\endgroup$ – Gopal Anantharaman Sep 17 '18 at 17:13
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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.

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