I'm sure this has a very "duh" answer, but it never occurred to me before and now it's really bothering me. Add it to the long list of things I never questioned about math until I started teaching it.

Consider the piece-wise function {x^2 if x<0; x^2 + 1 if x>=0}. The function is discontinuous at x=0, therefore the derivative does not exist. But the derivative is a limit, and a limit will exist as long as the right and left hand sides are equal. For this function, the left hand derivative is equal to the right hand derivative since both are zero, so doesn't this mean the limit and therefore the derivative exists? I'm trying to think of an argument based on real analysis, but it's been about 15 years since I took the class and I'm a little rusty. I need an explanation I can give to high school students.

## Simple calculus question that is bugging me -answered,thanks

### Simple calculus question that is bugging me -answered,thanks

Last edited by boo78 on Tue Oct 22, 2013 4:09 pm, edited 1 time in total.

### Re: Simple calculus question that is bugging me

Just look at the definition of the derivative.

$$f^\prime(x)=\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}.$$

$$\lim_{h \rightarrow 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \rightarrow 0^+}\frac{h^2+1 - 1}{h} = \lim_{h \rightarrow 0^+}h = 0,$$

but

$$\lim_{h \rightarrow 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \rightarrow 0^-} \frac{h^2 - 1}{h} = \lim_{h \rightarrow 0^-} h - \frac{1}{h} = \lim_{h \rightarrow 0^-} - \frac{1}{h} = +\infty,$$

so the derivative does not exist at $$x=0$$.

$$f^\prime(x)=\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}.$$

$$\lim_{h \rightarrow 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \rightarrow 0^+}\frac{h^2+1 - 1}{h} = \lim_{h \rightarrow 0^+}h = 0,$$

but

$$\lim_{h \rightarrow 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \rightarrow 0^-} \frac{h^2 - 1}{h} = \lim_{h \rightarrow 0^-} h - \frac{1}{h} = \lim_{h \rightarrow 0^-} - \frac{1}{h} = +\infty,$$

so the derivative does not exist at $$x=0$$.

### Re: Simple calculus question that is bugging me

Yeah, I was thinking it over just now on my way home and that came to me. I was thinking in terms of finding the derivative function, not the derivative at a specific point. I knew it didn't exist, but it's like those funky "find what's wrong with this" arguments, like pi=4. Like I said, it was a "duh", and it always seems with those that the answer occurs to me right after I ask. I think my brain is still a bit fried after the test Saturday.

### Re: Simple calculus question that is bugging me

For function have derivative at a point, it is necessary that the function must be continuous at that point. Your function is obviously not continuous at 0. So, it can't have a derivative at that point.

### Re: Simple calculus question that is bugging me -answered,thanks

Yeah, I knew it didn't exist. I was just having a brain-fart. What got me tripped up was that just because the limit of a limit exists doesn't mean the limit itself exists, if that makes any sense to anyone other than me. For those of you about to become TAs, a cautionary note: Nothing will make you feel like more of an idiot than trying to teach something you think you know really well. 6 years of teaching has taught me that lesson well.