How Is the Derivative of ln(x) Proven Using Limits?

[Jameson]

Define the derivative of the natural logarithm to be: [math]\frac{d}{dx} \ln(x) = \frac{1}{x}[/math]

Demonstrate this rule is valid by using the limit definition of a derivative.

Hint:
[sp]
This definition of the exponential function is necessary to calculate the limit.
[math]e^x = \lim_{n \to \infty} \left({1 + \frac x n}\right)^n[/math]
[/sp]

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[Jameson]

Congratulations to the following members for their correct solutions:

1) Sudharaka
2) checkittwice
3) Reckoner
4) veronica1999

Solution (from Reckoner):

[sp]\[\frac d{dx}\ln(x) = \lim_{\Delta x\to 0}\frac{\ln\left(x+\Delta x\right) - \ln(x)}{\Delta x}\]

\[=\lim_{\Delta x\to0}\frac1{\Delta x}\ln\left(\frac{x+\Delta x}x\right)\]

\[=\lim_{\Delta x\to0}\frac1{\Delta x}\ln\left(1+\frac{\Delta x}x\right)\]

\[=\lim_{\Delta x\to0}\ln\left(\left[1+\frac{\Delta x}x\right]^{1/\Delta x}\right)\]

Let \(n = \frac1{\Delta x}\). Then for \(\Delta x > 0, n\to\infty\) as \(\Delta x\to0\) and for \(\Delta x < 0, n\to-\infty\) as \(\Delta x\to0\). So we have

\[\lim_{\Delta x\to0^+}\ln\left(\left[1+\frac{\Delta x}x\right]^{1/\Delta x}\right) = \lim_{n\to\infty}\ln\left(\left[1+\frac{1/x}n\right]^n\right)\]

\[=\ln\left(e^{1/x}\right) = \frac1x.\]

Note that the continuity of \(\ln(x)\) allowed us to move the limit inside the function. Similarly,

\[\lim_{\Delta x\to0^-}\ln\left(\left[1+\frac{\Delta x}x\right]^{1/\Delta x}\right) = \lim_{n\to-\infty}\ln\left(\left[1+\frac{1/x}n\right]^n\right)\]

\[ = \lim_{n\to\infty}\ln\left(\left[1+\frac{1/x}{-n}\right]^{-n}\right)\]

\[ = \lim_{n\to\infty}-\ln\left(\left[1+\frac{-1/x}n\right]^n\right)\]

\[ = -\ln\left(e^{-1/x}\right) = \frac1x.\]

Since both one-sided limits exist and are equal to \(\frac1x\),

\[\lim_{\Delta x\to0}\ln\left(\left[1+\frac{\Delta x}x\right]^{1/\Delta x}\right) = \frac1x\]

and thus,

\[\frac d{dx}\ln(x) = \frac1x.\][/sp]