I don't get how the concept of limits allows us to move from secants to tangents. From what I've gathered, to find the derivative/tangent line at a point, we take a secant line, move the points infinitely close to each other, and find the slope value that it is getting infinitely close to. That slope value will be the limit, the derivative, the slope of the tangent line.

What I don't get is how we can make this logical leap, that because the limit as x approaches a is approaching a certain value, the value at a (the derivative) must be the value of a limit.

Because the same doesn't apply for limits. Let's take a function with a removable discontinuity for instance. enter image description here

Lim x->4=0. But at x=4, the value won't be 0! So how can we make the assumption that because the secant shrinks and gets closer and closer to being a tangent, lim x->a approaches a value, then at a, the derivative will be that value?

I have a gut feeling that the reason why it works out is because all differentiable functions are continuous. But can you elaborate and explain on this statement? Because I don't see how I can justify my intuition. Can someone someone prove/show to me that because all differentiable functions are continuous, when lim x->a approaches a value, when actually at that value, the derivative will be equals to that?

But then if continuity is the thing that makes it work, how come not all continuous functions are differentiable?

Can someone please explain the above to me mathematically? Is there a way to explain this, without epsilon delta proofs, to someone learning Khan Academy Calculus?

  • $\begingroup$ :all continuos functions are differentiable except the ones who have a bend, cusp, or vertical tangent. $\endgroup$ – Faraday Pathak Sep 4 '18 at 15:21
  • $\begingroup$ @veereshpandey Thanks, but can you answer the question, because that doesn't really help. I still don't see how limits allow the jump from secants to tangents. $\endgroup$ – user577730 Sep 4 '18 at 15:21
  • $\begingroup$ In the second paragraph, you say "...the value at $a$ (the derivative)...". This is vague. The value at $a$, through which function? The answer is not the original function, for example. $\endgroup$ – user373239 Sep 4 '18 at 15:44
  • $\begingroup$ This question is completely unclear to me. What do you mean by “the same doesn't apply for limits” and “Lim x->4=0”? The limit of what is zero?!? $\endgroup$ – Hans Lundmark Sep 4 '18 at 18:58

So, let's review the definition of a secant line, the definition of a tangent line, and the definition of the derivative:

A secant line to a curve is a line that intersects the curve in two distinct points.

The derivative of a function $f(x)$ at point $a$, denoted $f'(a),$ is defined to be the limit $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}, $$ provided this limit exists.

A tangent line to a curve $f(x)$ at a point $a$, if it exists, is defined to be that line with slope $f'(a)$ going through the point $(a, f(a)).$

(Side note: a tangent line is most definitely NOT a line that intersects a curve at only one place. This is a common, but misguided definition of the tangent line often encountered in high school geometry textbooks and is incorrect in general, though true for conic sections such as circles, ellipses, hyperbolas, and parabolas. It's better to think of tangent lines in terms of the derivative.)

First off, notice that the derivative is defined as a limit. So, for the derivative to exist, the limit has to exist. There are multiple ways a limit in general can fail to exist (right-hand and left-hand limits not equal, infinities, etc.), and there are additional ways that a derivative-type limit can fail to exist. For example, while this is not sufficient for the derivative to exist, the function must be continuous at $a$. That means $\lim_{x\to a}f(x)=f(a),$ a statement which says three non-trivial things: 1. The limit exists. 2. $f(a)$ is defined. 3. The limit equals the value of the function. Any of those can fail. In addition, you could have, say, a corner at $a$. The function $f(x)=|x|$ has a corner at $a=0$. While the function is continuous there, $$1=\lim_{x\to 0^{+}}\frac{|x|-|0|}{x-0}\not=\lim_{x\to 0^{-}}\frac{|x|-|0|}{x-0}=-1.$$ The function $f(x)=|x|$ is therefore not differentiable at $x=0,$ because the required limit does not exist. Geometrically, think about the corner this way: what unique value could you assign to the "slope" of $|x|$ at $x=0?$ You could draw any number of candidates. What the limit definition is saying is that you have to be able to come up with a unique value for the slope at a point, in order for the function to be differentiable at that point.

As for the relationship between secant lines and tangent lines, just notice that the slope of the secant line to $f$ at the points $x$ and $a$ is given by $$m=\frac{f(x)-f(a)}{x-a}.$$ The slope of the tangent line to $f(x)$ at $x=a$ is defined as the limit of the secant line slopes. You can imagine, geometrically, that you're letting the point $x$ get arbitrarily close to $a$. In the limit, you get the tangent line (provided the limit exists).


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