Is $f : (-1, 1) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$ differentiable at $x = 0$?

Question

Is $f : (-1, 1) \to \mathbb{R}$ defined by $$f(x) = \frac{1}{x}$$ differentiable at $a = 0$?

By the definition of the derivative I get the following $$f'(a) = \lim_{t \to 0} \frac{f(a+t) - f(a)}{t} = \lim_{t \to 0} \frac{\frac{1}{a+t}-\frac{1}{a}}{t}= \lim_{t \to 0} \frac{-1}{a^2 + at} = -\frac{1}{a^2}$$

so $f$ is differentiable at $a$. However if $f$ is differentiable at $a$ then $f$ must be continuous at $a = 0$ a contradiction since $f$ is not continuous at $a = 0$.

What have I done wrong here?

Answer 1

The function $f'(x)=\frac{-1}{x^2}$ is not defined at $x=0$ (uniquely there) then it isn't continuous and $f(x)$ isn't differentiable at $x=0$.

Answer 2

Look at what you were doing here: $$f'(a) = \lim_{h \to 0}\frac{f(a + h) - f(a)}{h}$$

When you wrote that, you need to remember that the $f$ you're talking about is $f(x) = \frac{1}{x}$, and that the $a$ you're talking about is $0$. Replacing those things, this is what you wrote: $$f'(0) = \lim_{h \to 0}\frac{\frac{1}{0 + h} - \frac{1}{0}}{h}$$

A limit, essentially, is a question: what happens to this number when I make $h$ small? In this case, "this number" doesn't exist - to find it, you'd need to divide by zero! So what happens to it is that it stays not existing, and $f'(0)$ therefore doesn't exist. And what does "not differentiable at $0$" mean if not "$f'(0)$ does not exist"?

Alternatively, you can see the issue from your conclusion, too. Remember, just because you can write something down doesn't mean it exists - I can write "$x + 1 = 2$ and $x + 2 = 4$", but that doesn't mean there is such a number. In this case, you found that $f'(a) = -\frac{1}{a^2}$ for any $a$; but only for $a \neq 0$ does this thing you've written exist.