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]

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    $\begingroup$ $$2=e^{\ln2}$$ and math.stackexchange.com/questions/152605/… $\endgroup$ – lab bhattacharjee Aug 30 '15 at 9:50
  • $\begingroup$ Use L'Hôpital's rule en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule $\endgroup$ – user261263 Aug 30 '15 at 9:55
  • $\begingroup$ @EugenCovaci using L'Hospital's rule here might be considered to be circular. $\endgroup$ – Hirshy Aug 30 '15 at 9:57
  • $\begingroup$ @Hirshy Can you explain why? $\endgroup$ – user261263 Aug 30 '15 at 9:58
  • $\begingroup$ @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: $\lim\limits_{x\to 0}\frac{\sin(x)}{x}$). $\endgroup$ – Hirshy Aug 30 '15 at 10:02

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).

  • $\begingroup$ 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. $\endgroup$ – Malcolm Aug 30 '15 at 10:45
  • $\begingroup$ 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. $\endgroup$ – Hirshy Aug 30 '15 at 10:54

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 $$\lim\limits_{x\rightarrow 0^-}{\frac{2^x-1}{x}} = \lim\limits_{x\rightarrow 0^+}{\frac{2^{-x}-1}{-x}} = \lim\limits_{x\rightarrow 0^+}{\frac{1}{2^x}\frac{2^x-1}{x}} = \lim\limits_{x\rightarrow 0^+}{\frac{2^x-1}{x}}. $$


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