pbuk
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Because you can prove it.NoahsArk said:However, what I don't know is why it is EXACTLY 2x and not just very close to 2x.
- Start by assuming the opposite: that the slope of the tangent line is not exactly ## 2x ##, it is ## 2x + \varepsilon ##.
- Now you already know that for ## f(x) = x^2 ## the slope of the secant line between ## x ## and ## x + \Delta x ## is ## 2x + \Delta x ##.
- As you make ## \Delta x ## smaller, the slope of the secant line will become closer to the slope of the tangent line - or rather it will never start getting further away (this is the definition of differentiability). So if you believe that the slope of the tangent is actually ## 2x + \varepsilon ## then the slope of the secant line can never be smaller than this.
- Choose ## \Delta x = \frac \varepsilon 2 ##. The slope of the secant line is ## 2 x + \Delta x = 2 x + \frac \varepsilon 2 ##. From 3. you know that the slope must be greater than or equal to ## 2 x + \varepsilon ##. The only value of ## \varepsilon ## for which ## 2 x + \frac \varepsilon 2 \ge 2 x + \varepsilon ## is ## 0 ##.
- From 1. the slope of the tangent line is ## 2x + \varepsilon = 2x + 0 = 2x ##.
@NoahsArk if you can spot the hole and it bothers you then you might find the book Calculus by Michael Spivak interesting. If you prefer learning from a combination of video lectures, notes and exercises then I recommend https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/.