# Verification of the proof that the derivative of 1/x exists using the definition.

I am trying to prove that the derivative of 1/x exists. Is my proof correct? Please help me!

Proof.

According to the definition of the derivative.

It is necessary to analyze two cases:

Case numer 1: x is not equal to 0.

For this case f(x), f(a), x, and a exist because 1/x is a continuous function everywhere except when x=0 because x is a polynomial and 1/x is undefined only when the denominator is equal to 0, what happens only when x=0.

Therefore the derivative exists when x is not equal to 0.

Case numer 2: x is equal to 0.

Let analyze the graph of 1/x.

1/x is discontinuous at x=0 because the left limit and right limit of 1/x is not the same.

As continuity is a necessary condition for differentiability, 1/x is not differentiable at x=0. Q.E.D.

• You've demonstrated that $f$ is continuous for $x\neq 0$; however, continuous does not imply differentiable. To show that it is differentiable, you need to show that the limit that defines the derivative exists at all points $x\neq 0$. – Nick Peterson Nov 2 '16 at 18:32
• Hint: Analyze $\lim_{x\rightarrow a}\frac{\frac{1}{x}-\frac{1}{a}}{x-a}$ and verify that the limit is indeed $\frac{-1}{x^2}$ (for $x\neq 0$). – Justin Benfield Nov 2 '16 at 18:34
• @NickPeterson Thank you for your help! I have a question. As f(x), f(a), x and a exist, then f(x)-f(a) and x-a exist. Then the limit and the derivative exist, or not? – Beginner Nov 2 '16 at 18:35
• @JustinBenfield Thank you for your help and you hint! May you read my last comment and answer that question? – Beginner Nov 2 '16 at 18:36
• You get $0/0$ when you attempt to directly evaluate the limit, this is what is known as an indeterminant form. This does not reveal whether or not the limit is actually defined (you need a determinant form) (ref: en.wikipedia.org/wiki/Indeterminate_form). – Justin Benfield Nov 2 '16 at 18:36

For the case $a \neq 0$ we have that
$$\lim_{x \to a} \frac{f(x) - f(a)}{x-a} = \lim_{x \to a} \frac{\frac 1 x - \frac 1 a}{x-a} = \lim_{x \to a} \frac{\frac{a - x}{ax}}{x-a} = \lim_{x \to a} -\frac{1}{ax} = -\frac{1}{a^2}.$$
If $a = 0$ then $f$ is not continuos and hence not differentiable.
• Should be $-1/a^2$ at the end, since $x\rightarrow a$, not other way around. – Justin Benfield Nov 2 '16 at 18:46