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Can someone help me find this limit here. I only know how to use L'Hospital's rule but I want to be able to evaluate this limit without using differentiation. $$\lim \limits_{h \to 0} \frac{\pi^h-1}{h}$$

The reason I want this limit is because just like $e$ can be expressed as $\sum_{n=0}^{\infty}\frac{1}{n!}$ I want to find a way to do the same with $\pi$ so i want to find the dervivative of $\pi^x$ without having $\pi$ in the result.

Ziad Fakhoury
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  • There is no way to avoid derivatives with this. The limit is $\lim_{h \to 0} \frac{\pi^h - \pi^0}{h-0}$, which is a difference quotient. – GFauxPas Jun 24 '16 at 12:48
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    If you are allowed to use the fact that $\lim_{h\to0} \frac{\mathrm{e}^h -1}{h}=1$, the you may use the substitution $t = h\log \pi$ to compute your limit. – Sangchul Lee Jun 24 '16 at 12:49
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    @GFauxPas However, $\lim_{h \to 0} \frac{e^h-1}{h-0}$ is also a difference quotient, but it can be calculated without derivatives. – Noble Mushtak Jun 24 '16 at 12:49
  • Do you allow expanding into Taylor series? Because expanding $\pi^h = e^{h \ln \pi}$ gives this answer easily. – sTertooy Jun 24 '16 at 12:49
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    @SteamyRoot That's still derivatives. – Noble Mushtak Jun 24 '16 at 12:50
  • Related question http://math.stackexchange.com/questions/1491220/show-lim-limits-h-to-0-fracah-1h-exists-without-lh%C3%B4pital-or-even-r – quid Jun 24 '16 at 12:51
  • That one may be still closer http://math.stackexchange.com/questions/331153/proving-that-lim-h-to-0-fracbh-1h-lnb I think it could be a duplicate; – quid Jun 24 '16 at 12:52
  • In your latest edit you added a reason that you want this limit. However, you are not going to learn anything about $\pi$ by this method. Why not? Because there is a completely general formula: $\lim_{h \to 0} \frac{a^h-1}{h} = \ln(a)$. Whatever method you find for proving this formula in the special case $a=\pi$ will work for any $a>0$. – Lee Mosher Jun 24 '16 at 13:06
  • @LeeMosher So really I'm looking for a way to express the natural log of a number rationally. Is that possible? – Ziad Fakhoury Jun 24 '16 at 13:10
  • I don't know what you mean by "express ... rationally". – Lee Mosher Jun 24 '16 at 13:23

3 Answers3

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There's no way we can avoid $\pi$ here. The value of this limit is $\log\pi$. Despite the method we use to get that limit, the exact value is $\log\pi$ and there's no way to avoid that. We can, however, do this without limits.

Let's say $\pi=e^{\ln \pi}$ $$\lim_{h \to 0} \frac{e^{h\ln\pi}-1}{h}$$ Now, we take @SangchiLee's advice and say $t=h\log \pi$, so that $h=\frac{t}{\log \pi}$: $$\lim_{t \to 0} \frac{e^t-1}{\frac{t}{\log \pi}}$$ Simplify: $$\lim_{t \to 0} \log\pi\frac{e^t-1}{t}$$ Take the $\log\pi$ out of the limit: $$\log\pi\lim_{t \to 0} \frac{e^t-1}{t}$$ Now, the limit on the right is equal to $1$, so we have: $$\log\pi\cdot 1=\log\pi$$

Noble Mushtak
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$$\lim_{x\to 0} \frac{e^{hln\pi}-1}{h} = (H) = $$ $$\lim_{x\to 0} \frac{ln{\pi} * lne^{hln\pi}}{1} = ln \pi $$

plktrautman
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You don't need that limit to compute the derivative of $a^x$ though. ($a$ a constant)

Since $a = e^{ \ln a}$ (log and exponential are inverse functions), you have $a^x = e^{(\ln a )x}$, so $$ \frac{d}{dx}(a^x) = \frac{d}{dx} e^{(\ln a)x} = (\ln a) e^{(\ln a) x}$$

D_S
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