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Edit: From the comment below it seems like the question behind is:

How can we determine whether of not the function $f(x) = x\sin(x) / x$ has a tangent at $x=0$. My thought is that one would have to find $$ \lim_{x\to 0} \frac{x\sin(1/x) - 0\sin(1/0)}{x - 0} $$ and I think this might be equal to $$ \sin(1/x). $$ I'm just trying to show that the tangent at $x = 0$ for $x\sin (1/x)$ does not exist, by showing that there is no limit for the "gradient graph" as $x$ tends to $0$.

Thomas
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Russ_Student
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    what is $1/0 =$? – Prahlad Vaidyanathan Oct 04 '13 at 17:29
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    $\frac{1}{0} $ is not defined. – copper.hat Oct 04 '13 at 17:30
  • $ \sin(y) \in [-1,1]$ for any real $y$ and so $ 0 \sin(y) = 0$ for any real $y$. Unfortunately $\frac10$ is not real. – Henry Oct 04 '13 at 17:37
  • Even if you do ignore the $0\sin(1/0)$, you still have the issue that $\sin(1/x)$ on the right hand side does not converge as $x \to 0$ so it may not work as a limit as you have written it. Perhaps you mean the limit of the difference between the left and the right hand sides tends to $0$. – Henry Oct 04 '13 at 17:42
  • I'm just trying to show that the tangent at x = 0 for x sin (1/x) does not exist, by showing that there is no limit for the "gradient graph" as x tends to 0. – Russ_Student Oct 04 '13 at 17:50
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    I think the OP is trying to calculate the derivative of $x\sin(\frac{1}{x})$ at $0$ (if well-defined) from the definition but has messed up the limit. – Dan Rust Oct 04 '13 at 18:13
  • @user2000304: I edited your question and I tried to get at what you asking. Please review the question and check that this is what you want to ask. – Thomas Oct 04 '13 at 18:15

2 Answers2

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I would like to add on to Thomas' answer that, even if we extend the function to $0$ by setting $f(0)=\lim_{x\to 0}f(x)$, and so in our case $f(0)=0$ as $\lim_{x\to 0}x\sin(\frac{1}{x})=0$, the derivative does not exist.

This is because the derivative exists if and only if the limit $$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$ exists, and this is equal to $\lim_{h\to 0}\sin(\frac{1}{h})$ which does not exist.

Dan Rust
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The function $$ f(x) = x\sin(1/x) $$ is not defined at $x=0$. That is $f(0)$ is not defined. So, no, there is not tangent at $x=0$ simply because $f$ is not defined at $0$ and so the there is no derivative at $x = 0$.

This you see, because when you are trying to use the definition of the derivative to find this at $0$, then you are having to evaluate $\sin(1/0)$, which is not defined. This might explain the confusion.

Thomas
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  • Hi thomas, thanks for answering. But why is there a tangent for x^2 sin (1/x)? (at x = 0, my lecturer tells us that there is a tangent) – Russ_Student Oct 04 '13 at 18:29
  • @user2000304: There isn't a tangent for $x^2\sin(1/x)$ at $x=0$. However, if you defined $f(0) = 0$ (as hinted at by Daniel in his answer) then this function would actually be differentiable at $0$ with derivative $\lim_{x\to 0} x\sin(1/x) = 0$. – Thomas Oct 04 '13 at 18:35