Analyzing the continuity and derivability of $\frac{1-\cos(x)}{x}$ at 0

Question

There is a piecewise function $f$ defined as following:

$$f:\left\{\begin{matrix} \frac{1-\cos(x)}{x} \Leftarrow x\neq0\\ 0 \Leftarrow x=0 \end{matrix}\right.$$

I'm asked to first show that $f$ is continuous at $0$, then show that $f$ is derivable at $0$, and finally calculate $f'(0)$.

I know that $g(x)$ is continuous at $x_{0}$ if $\lim_{x\rightarrow x_{0}}g(x)=g(x_{0})$ and derivable at $x_{0}$ if $\lim_{x\rightarrow x_{0}}\frac{g(x)-g(x_{0})}{x-x_{0}}=g'(x_{0})$ exists.

Here's what I have done to demonstrate that $f$ is continuous at $0$:

$$\lim_{x\rightarrow 0}\frac{1-\cos(x)}{x}=\lim_{x\rightarrow 0}-\frac{\cos(x)-\cos(0)}{x-0}=-\lim_{x\rightarrow 0}\frac{\cos(x)-\cos(0)}{x-0}=-\cos'(0)=\sin(0)=0=f(0)$$

Then, I was confused when I saw the following question that asked me to show that $f$ is derivable at $0$, as that property was how I solved the first question.

I think that there should be some another way to demonstrate how $f$ is continuous at $0$ which I don't realize, or it could be that I'm misusing the conditions of continuity and derivability at a point.

Anyways, any help would be appreciated and thank you for your time.

Answer 1

You did well for establishing continuity at $0$.

Every time a function $g$ (defined over an open interval $(a,b)$, for simplicity) is differentiable at a point $x_0\in(a,b)$, the function $$ f(x)=\begin{cases} \dfrac{g(x)-g(x_0)}{x-x_0} & \text{if $x\in(a,b)$, $x\ne x_0$} \\[6px] g'(x_0) & \text{if $x=x_0$} \end{cases} $$ is continuous at $x_0$, by definition of derivative and of limit. In your case, $g(x)=-\cos x$, $x_0=0$ and $g'(0)=\sin 0=0$.

For differentiability at $x_0$ you can apply the definition: $$ f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0} $$ if the limit exists and is finite. In your case, $$ \lim_{x\to0}\frac{\dfrac{1-\cos x}{x}-0}{x}= \lim_{x\to0}\frac{1-\cos x}{x^2}= \lim_{x\to0}\frac{1-\cos^2x}{x^2}\frac{1}{1+\cos x}= \frac{1}{2} $$ The first factor in the final limit is $$ \left(\frac{\sin x}{x}\right)^2 $$

Answer 2

You can do it using only basic methods. First let $x = 2y$, then use trigonometric equality:

$$ 1 - \cos(x) = 1 - \cos(2y) = 1 - (1 - 2 \sin^2(y)) = 2 \sin^2(y) $$

As you should know $\lim_{x\to 0}\frac{\sin x}{x} = 1$, so here you have

$$ \frac{2 \sin^2{y}}{2y} = \frac{\sin(y)}{y} \cdot \sin(y) \to 1 \cdot 0 = 0 $$

Now you can use definition:

$$ \lim_{x \to 0}\frac{f(x)- f(0)}{x} = \lim \frac{(1 - \cos(x))- 0}{x^2} = \lim \frac{2\sin^2(y)}{4y^2} = \frac{1}{2} \cdot 1^2 = \frac{1}{2} $$

However, your solution is quite correct. But if you didn't proved that cosinus is differentiable, then probably you should do it first!

Answer 3

Continuity: Note that

$$\lim \limits_{x \to 0^-} f(x)=\lim \limits_{x \to 0^+} f(x)=\lim \limits_{x \to 0}\frac{1-\cos x}{x}$$

Apply L' rule, $$=\lim \limits_{x \to 0}\frac{\sin x}{1}=0=f(0)$$

So $f$ is continuous at $x=0$.

Differentiability: Note that

$$\lim \limits_{x \to 0^-} \frac{f(x)-f(0)}{x-0}=\lim \limits_{x \to 0^+} \frac{f(x)-f(0)}{x-0}=\lim \limits_{x \to 0}\frac{1-\cos x-0}{x^2}$$

Apply L' rule,

$$=\lim \limits_{x \to 0}\frac{\sin x}{2x}=1/2$$ So, $$f'(0)=1/2$$