# Is the Function B(x)= xsin(1/x) Differentiable at x=0?

• sl2382
In summary, the function B(x) is defined as xsin(1/x) when x is not equal to 0 and 0 when x is equal to 0. The question is whether the function is differentiable at 0. After attempting the solution, it is determined that the function is not differentiable at 0 because the derivative does not exist at that point. However, the function is defined at 0, so it is a piecewise function. To be differentiable, the function must meet the requirements of having the derivative exist at a point and the limits approaching that point from both sides being the same. Using the definition of the derivative as a difference quotient, it is shown that the limit does not exist. A
sl2382

## Homework Statement

B(x)= xsin(1/x) when x is not equal to 0

= 0 when x is equal to 0

Determine if the function is differentiable at 0

## The Attempt at a Solution

I get B'(x)= sin(1/x)+cos(1/x)*(-1/x) but really do not know what should be done next.. I think for B'(x) x cannot be 0, but isn't the discontinuity removed as the function is defined to be 0 at x=0? ...

So it's a piecewise function right?

What does the function have to be in order for it to be differentiable? Check with the definition of the derivative.

QuarkCharmer said:
So it's a piecewise function right?

What does the function have to be in order for it to be differentiable? Check with the definition of the derivative.

I think the requirements are: 1)the derivative exists at a point 2)limits approaching from both sides of that point are the same ?

So, the derivative does not exist at 0, BUT isn't it defined at 0 ? Does that mean the derivative actually exists and the function can be differentiate?

Write the derivative as a difference quotient. f'(0) should be lim h->0 (f(h)-f(0))/h. Pick a specific sequence approaching 0, say h_n=1/(pi*n/2) for n an integer. So h_n->0 as n->infinity. Is there a limit? It's actually pretty helpful to sketch a graph of the function.

## 1. What does it mean for a function to be differentiable at x=0?

Being differentiable at x=0 means that the function has a well-defined derivative at that point. In other words, the function is smooth and continuous at x=0, and there is no sudden change or discontinuity in the slope of the function at that point.

## 2. How do you determine if a function is differentiable at x=0?

A function is differentiable at x=0 if the limit of the difference quotient, also known as the derivative, exists at that point. This means that the function must be continuous at x=0 and the left and right-hand limits must be equal.

## 3. Can a function be continuous but not differentiable at x=0?

Yes, a function can be continuous but not differentiable at x=0. This can happen if the function has a sharp corner or cusp at that point. In such cases, the left and right-hand limits of the difference quotient will not be equal, and therefore the derivative does not exist.

## 4. What is the difference between differentiability and continuity at x=0?

Continuity at x=0 means that the function is smooth and there are no sudden jumps or breaks in the graph at that point. Differentiability at x=0 means that the function is not only continuous, but also has a well-defined derivative at that point. In other words, differentiability implies continuity, but continuity does not necessarily imply differentiability.

## 5. Can a function be differentiable at x=0 but not continuous?

No, a function cannot be differentiable at x=0 if it is not continuous. This is because for a function to be differentiable, it must also be continuous and have equal left and right-hand limits at that point. If a function is not continuous at x=0, it cannot have a well-defined derivative at that point.

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