Is there a tangent of $x\sin(1/x)$ at $x = 0$?
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 $0$.
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
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