# A function's derivative being not defined for some X but having a limit

• B
• Mcp
In summary, the function is differentiable at x=0 if the limits of its derivative exist and are equal. The existence of a closed form representation is irrelevant. A function cannot be differentiable at points where it is not defined. However, if the function is defined in a differentiable way at x=0, then it will be differentiable and the derivative at x=0 will be 0.
Mcp
Let's say I have a function whose derivative is (tan(x)-sin(x))/x. It is not defined for X=0 but as X approaches 0 the derivative approaches 0, so should I conclude that my function is not differentiable at X=0 or should I conclude that the derivative at X=0 is 0.

The function is differentiable at ##x=0## if the limits ##\lim_{h \to \pm 0} \dfrac{f(x+h)-f(x)}{h}## exist and are equal. That you cannot find a closed form which reflects this existence is irrelevant.

A function cannot be differentiable at points where it is not defined.

However if you define $$f(x) = \begin{cases} \lim_{\epsilon \to 0}\int_\epsilon^x \frac{\tan u - \sin u}{u}\,du, & x \neq 0, \\ 0, & x = 0\end{cases}$$ then $f$ will be differentiable and $$f'(0) = \lim_{x \to 0} \frac{f(x)}{x} = 0.$$

fresh_42
pasmith said:
A function cannot be differentiable at points where it is not defined.

However if you define $$f(x) = \begin{cases} \lim_{\epsilon \to 0}\int_\epsilon^x \frac{\tan u - \sin u}{u}\,du, & x \neq 0, \\ 0, & x = 0\end{cases}$$ then $f$ will be differentiable and $$f'(0) = \lim_{x \to 0} \frac{f(x)}{x} = 0.$$
Are you saying that derivative not being defined at X=0 also implies that the function is not defined at X=0?

Mcp said:
Are you saying that derivative not being defined at X=0 also implies that the function is not defined at X=0?
No. The other way around. No function defined, no derivative. If the function is defined and the derivative (as limit) exists, then it is differentiable, regardless of there is a closed expression or not. In your case we have
$$f'(x) = \begin{cases} \dfrac{\tan(x)-\sin(x)}{x} & \text{ if }x\neq 0\\ 0 & \text{ if } x=0 \end{cases}$$
The function ##f(x)## whose derivative is ##f'(x)## has to be defined at ##x=0## in an appropriate, i.e. differentiable way. If ##f(x)## isn't defined at ##x=0##, then the question whether ##\left. \dfrac{d}{dx}\right|_{x=0}f(x)## exists or not doesn't make sense.

Mcp

## 1. What does it mean for a function's derivative to be not defined for some X?

When a function's derivative is not defined for some X, it means that the function is not differentiable at that particular point. This means that the slope of the tangent line at that point does not exist, and the function has a sharp change or discontinuity at that point.

## 2. Can a function have a limit but not be differentiable at that point?

Yes, it is possible for a function to have a limit at a certain point but not be differentiable at that point. This occurs when the function has a sharp change or discontinuity at that point, making it not differentiable. However, the limit of the function at that point can still exist.

## 3. How can we determine if a function is differentiable at a point?

A function is differentiable at a point if the limit of the difference quotient, also known as the derivative, exists at that point. This means that the slope of the tangent line at that point can be determined and the function is smooth at that point without any sharp changes or discontinuities.

## 4. What is the significance of a function having a derivative but not being defined at a point?

When a function has a derivative but is not defined at a point, it means that the function is not continuous at that point. This can have implications in real-world applications, as it indicates a sharp change or discontinuity in the behavior of the function at that point.

## 5. Can a function be differentiable at a point but not have a limit at that point?

No, a function cannot be differentiable at a point if it does not have a limit at that point. This is because the limit of the difference quotient, which is the derivative, is required for a function to be differentiable at a point. Therefore, if the limit does not exist, the function cannot be differentiable at that point.

• Calculus
Replies
1
Views
544
• Calculus
Replies
2
Views
670
• Calculus
Replies
9
Views
1K
• Calculus
Replies
4
Views
840
• Calculus
Replies
2
Views
672
• Calculus
Replies
3
Views
2K
• Calculus
Replies
3
Views
2K
• Calculus
Replies
5
Views
1K
• Calculus
Replies
1
Views
2K
• Calculus
Replies
6
Views
2K