# Is there any easy way to teach the concept of limit for differentiability notion for student in high school level?

Really am interesting to get easy and good example to teach the concept of limit for differentiability. courses for student in high school , I have used the notion of limit as calculus of images of functions , for example :$$f(x)=x^2+1, \lim_{x\to 0} (x^2+1)=f(0)=1$$ but am afraid they will take in their mind that all functions have limit at $$0$$, and in the same time this method can't give anything to differentiability concept , I have used again the physics notion, passage from middle velocity to velocity instantaneous of an object in motion \, but they haven't any idea about instantaneous velocity, then my question here is: How to teach them the concept of limit for getting differentiability notion at points using easiest way?

• matheducators.se might be more helpful for this issue.
– J.G.
Oct 5, 2019 at 20:26

There are two things you've said you want to tackle here. I'll start with the limit-at-$$0$$ issue, because it's very important to explain limits before derivatives (after all, derivatives are limits.) In fact, you should probably explain one-sided limits, then mention a "limit" exists when the two are equal.
Firstly, you're worried students will think there's always a limit at $$0$$, so use $$1/x$$ as a counterexample. We can't even say the limit is "infinity", because the right-hand limit is $$\infty$$, whereas the left-hand one is $$-\infty$$. And explain the only reason you've considered $$\lim_{x\to0}f(x)$$ in other examples is for convenience, because $$\lim_{x\to c}f(x)=\lim_{x\to0}f(x+c)$$ if it exists. (You can even show horizontal translation of the graph of a function to show how this works, so that $$\lim_{x\to1}\frac1x=\lim_{x\to0}\frac{1}{x+1}=1$$ is fine.)
For derivatives, you can explain that, while we call a function $$f$$ continuous at $$a$$ if $$f(a)=\lim_{x\to a}f(x)$$ or equivalently $$f(x)-f(a)\stackrel{x\to a}{\rightarrow}0$$, if $$f$$ is continuous at $$a$$ that doesn't say how quickly $$f(x)-f(a)$$ shrinks as $$x-a$$ shrinks. This allows you to explain that in some cases $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$, or perhaps more conveniently $$\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$, exists, but sometimes it doesn't. (You can explain why we care about such a limit by drawing secants and tangents, and using your usual examples.)
For instance, it does if $$f(x)=x^2$$ because $$\lim_{h\to0}\frac{(a+h)^2-a^2}{h}=\lim_{h\to0}(2a+h)=2a$$, but not if $$f(x)=|x|,\,a=0$$, because $$\frac{|h|}{h}$$ approaches different values either side of $$0$$. At least in that case the right- and left-hand limits exist. But if $$f(x)=\sqrt{x}$$ and $$a=0$$, there is no left-hand behaviour for $$f$$ (unless you bring in imaginary numbers - don't do that if they don't know them!), whereas on the right the limit would be $$\lim_{h\to0^+}\frac{1}{\sqrt{h}}=\infty$$. The upshot of all this is that one- or even two-sided continuity doesn't guarantee even one-sided differentiability.