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I would like to somehow firstly show that $$\lim_{x \to 0} \frac {2^x-1} x$$ exists and determine the value of the limit.

My first ideas were by Monotone Convergence. I have been able to prove that if $$a_x = \frac {2^x-1} x, \forall x \in \mathbb{R},$$ then $a_x>0, \forall x \in \mathbb{R}$. But I can't seem to show that it is a strictly decreasing sequence.

Furthermore, is there some way to determine the value of the limit?

[N.B: Those of you with a keen eye will notice that this question, in reality, is a part of the differentiating of $2^x$ by first principles. As a result, the limit must then be $\ln 2$. But for the purposes of exploring the method, I would like you to ignore that fact]

Hirshy
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Malcolm
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    $$2=e^{\ln2}$$ and http://math.stackexchange.com/questions/152605/proving-that-lim-limits-x-to-0-fracex-1x-1 – lab bhattacharjee Aug 30 '15 at 09:50
  • Use L'Hôpital's rule https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule –  Aug 30 '15 at 09:55
  • @EugenCovaci using L'Hospital's rule here might be considered to be circular. – Hirshy Aug 30 '15 at 09:57
  • @Hirshy Can you explain why? –  Aug 30 '15 at 09:58
  • @EugenCovaci to apply L'Hospital you need to know that the function $x\mapsto 2^x$ is differentiable whereas this limit that needs to be calculated is the definition of the derivative of $x\mapsto 2^x$ at $x=0$ (see my answer below). Thus you would calculate the derivative using a rule that first needs to know that the derivative exists. I wouldn't mind using L'Hospital to calculate this limit (depending on how you defined the exponential function), but I know some people would mind and not accept using L'Hospital (another example where this occurs: $\lim\limits_{x\to 0}\frac{\sin(x)}{x}$). – Hirshy Aug 30 '15 at 10:02
  • @Hirshy well, you don't actually need to calculate the deivative of $2^x$ from first principles. You could just re-write it as $e^{\ln(2)x}$ and use the chain rule. – Cataline Aug 30 '15 at 10:04
  • @Hirshy To apply L'Hospital I only need to prove the function $x↦2^x$ is differentiable in a vecinity of 0, except for 0. That's why there is nothing circular. Anyway, your answer is very good. –  Aug 30 '15 at 10:05
  • @Cataline yes, and I personally wouldn't mind using L'Hospital in this case. But as I said, I know some people who would be almost personally offended by this use of L'Hospital and just in case that OP is working with/for such a person, I wanted to point that out. In addition, I personally think that L'Hospital's rule is used too often to calculate these easy limits and therefor I try to avoid using it if possible. – Hirshy Aug 30 '15 at 10:08

2 Answers2

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Hint: write $$\lim\limits_{x\to 0}\frac{2^x-1}{x}=\lim\limits_{x\to 0}\frac{2^x-2^0}{x-0}.$$ This should hopefully look familiar to you (if not: think of the definition of the derivative).

Hirshy
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  • That would be the slope of $2^x$ at $x=0$

    I still don't quite see how that helps in determining the limit as you still need to somehow know what $\frac {d} {dx} 2^x$ at $x=0$ actually is.

     – Malcolm Aug 30 '15 at 10:45
  • Yes, you need to know the derivative of $x\mapsto 2^x$. If you don't know it yet, you can easily get it via $2^x=e^{x\cdot \ln(2)}$ and applying the chain rule. – Hirshy Aug 30 '15 at 10:54
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You can show that $x\mapsto\frac{2^x-1}{x}$ is an increasing function, which would imply that the left and right-hand limits exist at $x=0$. To do that, it suffices to show that $x\mapsto 2^x$ is convex. Since the mapping is continuous, it suffices to show midpoint convexity, i.e. $$2^{\frac{x_1+x_2}{2}}\le \frac{2^{x_1}+2^{x_2}}{2} $$ which is true by AM-GM. Now, to show that the left and right-hand limits are the same, we just calculate $$\lim\limits_{x\rightarrow 0^-}{\frac{2^x-1}{x}} = \lim\limits_{x\rightarrow 0^+}{\frac{2^{-x}-1}{-x}} = \lim\limits_{x\rightarrow 0^+}{\frac{1}{2^x}\frac{2^x-1}{x}} = \lim\limits_{x\rightarrow 0^+}{\frac{2^x-1}{x}}. $$

Joey Zou
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