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I recently made this post where the takeaway for me is that when evaluating a limit, the value at the actual point a does not matter, but only the values near a actually matter. That's why we can take an equation like:

$$\lim_{x \to 1} \frac{x^2-1}{x-1}$$

and simplify it to:

$$\lim_{x \to 1} \frac{(x+1)(x-1)}{x-1} = \lim_{x \to 1} x+1 = 2$$

So when evaluating the definition of a derivative, it seems like a similar concept is being applied right? Here is (one of) the definitions of a derivative:

$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$

where $a$ is a constant and $h$ is some value closer and closer to $a$.

Say we have a problem like:

Find an equation of the tangent line to the parabola $y = x^2$ at the point 1,1.

Solution

So $a = 1$

The slope therefore at 1 is:

\begin{align} \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h}&=\lim_{h \to 0} \frac{(1+h)^2 - 1}{h}\\ &=\lim_{h \to 0} \frac{1 + 2h + h^2 - 1}{h}\\ &=\lim_{h \to 0} \frac{2h + h^2}{h}\\ &=\lim_{h \to 0} 2+h \\ & = 2 \end{align} Are these two the same concepts? I'm just trying to tie together some points in my head.

And again, what we're doing is simplifying away the factor that leads to an indeterminate fraction right? And we can do this because we're not actually evaluating the fraction at $0$ but close to $0$? Is that right? Can anyone make this more clear?

Jwan622
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2 Answers2

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You are correct that the two ideas are closely linked. Once you define the tangent as the limit you cite, you evaluate the limit like any other. You don't have to evaluate the function at the point of interest, just a points nearby. When you say "Here is (one of) the definitions of a limit" that is not correct. This is a definition of the derivative. It is a definition that uses a limit. Once you get to the limit you evaluate it using the tools you have for evaluating limits.

Ross Millikan
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  • I guess my main question is... why does simplifying eliminate the indeterminate fraction and leave you an equation where you simply plug in the value that x is approaching? – Jwan622 Jan 08 '18 at 04:04
  • When you see an indeterminate form like $\infty - \infty$ it could be that one $\infty$ dominates over the other, like $\lim_{x \to \infty}(x^2-x)$. If you realize that it is easy, the limit is $+\infty$. It could be that the infinities are the same order, so you can remove them analytically and leave the rest, like $\lim_{x \to \infty} ((x+1)-x)$ Here you can subtract the $x$ terms analytically and have only a finite part left. – Ross Millikan Jan 08 '18 at 05:41
  • realize that it is easy? – Jwan622 Jan 08 '18 at 05:46
  • Sometimes it is easy, sometimes not. It depends on the problem how easy it is to see and do. – Ross Millikan Jan 08 '18 at 05:50
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Given some f(x), f(17) may or may not be determinate, but if all other f(x) between f(17 - 1 trillionth) & f(17 + 1 trillionth) are defined (an infinite number of them), it's useful to be able to arrive at the value of f, that gets better and better (the amount of error decreasing) the closer x gets to 17. That's limits.

Derivatives are different. They tell a story about a function's instantaneous slope: it's powerful to be able to write a function f' that spits out the slope of f at every valid value of x.

That the def'n of deriv. uses the limit is not the "big idea" of derivatives, even though limits are themselves a big idea. The "big idea" of the derivative is that it's a function's rise-to-run ratio, with both rise and run allowed to become obscenely teensy ("infinitesimal").

Neil Schipper
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