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Here's a question from my calculus textbook:

Show that when $\Delta x \to 0$, the increment in the function $y = 2^x$, corresponding to an increment $\Delta x$ in $x$, is, for any $x$, equivalent to the expression $2^x \ln 2\, \Delta x$.

We have that$$\Delta y = 2^{x + \Delta x} - 2^x = 2^x(2^{\Delta x} - 1).$$How do I show that as $\Delta x\to 0$, that $2^{\Delta x} - 1 \to \ln 2\,\Delta x$?

EDIT: Charles Hudgins in the comments says what we want to prove is a theorem that is usually merely cited. But I'm wondering if anyone can show this from first principles.

user1013124
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    The existence of the limit is a theorem (one you shouldn't have to prove directly in a calc class. Requires a result on convex functions). But that it is equal to the claimed value is essentially a definition. The work you need to show depends on how you have defined things. – Charles Hudgins Mar 02 '23 at 01:03
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    This is a hop, skip, and jump away from the definition of $e^x$ as a function. – Cameron Williams Mar 02 '23 at 01:11
  • $2^{\Delta x} - 1 \to \ln 2,\Delta x$ makes little sense to me, since when computing a limit $\Delta x\to 0$, the variable $\Delta x$ should not appear in the result. What would make sense is to show that $\lim_{\Delta x\to 0}\frac{2^{\Delta x}-1}{\Delta x}=\ln(2)$, which is the same as showing that the derivative of $f(x)=2^x$ at $0$ is $\ln 2$. Since $2^x=e^{x\ln(2)}$, this is straightforward if you know the derivative of the exponential function and the Chain Rule. – Taladris Mar 02 '23 at 01:21
  • Does this answer your question? Prove $\lim \limits_{h \to 0} \frac{a^h-1}{h} = \ln (a)$ without using L'hospital's. – QC_QAOA Mar 02 '23 at 02:34

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Should be careful how grouping is done.

$$\frac{dy}{dx}=y'=(2^x)'=2^x \ln 2\to\frac{\Delta y}{\Delta x}=2^x \ln 2 $$

Narasimham
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