Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The function $f$ is defined by $f(x)= \sin(1/x)$ for any $x\neq 0$. For $x=0$, $f(x)=0$. Determine if the function is differentiable at $x=0$.

I know that it isn't differentiable at that point because $f$ is not continuous at $x=0$, but I need to prove it and I'm not sure how to use

$$m(a)= \lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$

with a piecewise function.

share|improve this question
1  
LaTeX tip: Most function names are supposed to be upright, and you can get the common ones by preceding it with a backslash, e.g. \sin(x) instead of sin(x). If it is not built-in, you can use the \mathrm command, e.g. \mathrm{lcm}(a,b). You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. –  Zev Chonoles Mar 8 '13 at 17:22
4  
So just prove that it is not continous at $0$. –  1015 Mar 8 '13 at 17:22
    
or consider the difference quotient for $a=0$ at $x=\frac1{(2n+\frac12)\pi}$ –  Hagen von Eitzen Mar 8 '13 at 17:26

3 Answers 3

To show that the function is not differentiable at $0$, you need to show that the limit $$ \lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=\lim_{x \to 0}\frac{\sin(\frac 1x)}{x} $$ does not exist.

This can be done by finding two sequences $x_n$ and $y_n$ that both go to zero, but such that $$ \frac{\sin (\frac {1}{x_n})}{x_n} \text{ and } \frac{\sin (\frac {1}{y_n})}{y_n} $$ have different limits as $n \to \infty$.

As Hagen von Eitzen mention in his comment, trying reciprocals of multiples of $\pi$ should help you find appropriate sequences $x_n$ and $y_n$.

share|improve this answer

You need to prove that differentiability implies continuity.
Hint: consider the following limit
$$\lim_{x\to a}(f(x)-f(a))$$ and try to prove that it's equal to $0$ assuming that function $f$ is differentiable in point $a$, which means $f'(a)$ exists.

share|improve this answer

My approach: Consider the left and right limits.
i.e. right limit when x --> 0+ and left limit when x --> 0-.

Also, we may consider y = 1/x, and somehow "convert" the limit when x --> 0+ to become the limit when y --> infinity. Similarly, "convert" the limit when x --> 0- to the limit when y --> -infinity. These two limits should be different.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.