If function $f$ is differentiable at a point $a$, does it imply that $f'(a)$ exists?

Question

If function $f$ is differentiable at a point $a$, does it imply that $f'(a)$ exists?

I want to proof that if function $f$ is differentiable at a point $a$, then the function is continious at this point. I want to do this by using the multiplication rule for limits. For that I need to show that function $f$ has a finite limit at point $a$. I have read that function if is differentiable at a point $a$ if $f'(a)$ exists. However, it seems that it is not exactly what I need.

Thank you for your asnwers. I was a bit confused by how the definiton phrased.

I have an additional question: how does one know when "if" means "if and only if" and "if" means "if"?

Answer 1

The definition of "$f$ is differentiable at a point $a$" is that "$f'(a)$ exists". So $f$ is differentiable at $a$ if and only if $f'(a)$ exists.

(When you read that "function $f$ is differentiable at a point $a$ if $f'(a)$ exists", that was probably using the standard convention in mathematics that "if" actually means "if and only if" when making a definition. That is, if you write "I define $f$ to be differentiable at $a$ if $f'(a)$ exists", you actually mean that $f$ is differentiable at $a$ if and only if $f'(a)$ exists.)

Answer 2

Yes, it implies that $f'(a)$ exists and $f(x)$ is continuous at $x=a$. That's the definition of differentiability.

Answer 3

To say $f$ is differentiable at $a$ means that $f'(a) = \lim\limits_{h\to0} \dfrac{f(a+h) - f(a)} h$ exists. That is a definition, not something requiring proof.

If you want to prove that if $f$ is differentiable at $a$ then $f$ is continuous at $a$, you can do so as follows.

First, let us note that trivially $\lim\limits_{h\to0} \dfrac{f(a+h) - f(a)} h$ is the same as $\lim\limits_{x\to a} \dfrac{f(x)-f(a)}{x-a}$. That that limit exists is another way of saying what it means to say $f$ is differentiable at $a$.

Since $\lim\limits_{x\to a} \dfrac{f(x)-f(a)}{x-a} \vphantom{\dfrac1{\displaystyle\sum}}$ exists (and that means it's a finite number, not $+\infty$ or $-\infty$) and $\lim\limits_{x\to a} (x-a)$ exists, the limit of the product exists and is equal to the product of the two limits: $$ \lim_{x\to a} \left(\frac{f(x)-f(a)}{x-a} \cdot (x-a)\right) = \left(\lim_{x\to a} \frac{f(x)-f(a)}{x-a}\right) \cdot \left(\lim_{x\to a} (x-a) \right). $$ The first limit is some finite number (since $f$ is differentiable at $a$) and the second is $0$. Therefore $$ \lim_{x\to a} (f(x)-f(a)) = 0. $$ And given that, one can show that $$ \lim_{x\to a} f(x) = f(a), $$ so $f$ is continuous at $a$.