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I can looking for a simple example to illustrate $\lim\limits_{x \to x_0} xf(x) \neq x_0 \lim\limits_{x \to x_0} f(x)$

For example I have tried $f(x) = x-1, x_0 = 1$ hoping that I would get a zero on one side and a non-zero on the other, but so far without success.

Can someone provide an example to this statement?

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    $\begingroup$ $\lim_{x \to 0} x\frac{1}{x}$, but if you insist that $\lim_{x \to x_o} f(x)$ needs to always be defined, then such an example doesn't exist $\endgroup$ – AlkaKadri Nov 28 '18 at 0:45
  • $\begingroup$ Take $f(x) = \frac{1}{x}$ and $x_0 = 0$. Then $$\lim_{x\to x_0} x f(x) = \lim_{x\to 0} 1 = 1, $$ but $x_0 \lim_{x\to x_0} f(x)$ is not defined. $\endgroup$ – Xander Henderson Nov 28 '18 at 0:45
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HINT: If $\lim_{x\to x_0}f(x)$ exists then by the product rule for limits $$\lim_{x\to x_0}xf(x) =\left(\lim_{x\to x_0}x\right)\left(\lim_{x\to x_0}f(x)\right) =x_0\lim_{x\to x_0}f(x),$$ so you want to find some function $f$ and some point $x_0$ such that $\lim_{x\to x_0}f(x)$ does not exist.

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    $\begingroup$ Existence of the limit is enough for the product rule; continuity is not required. $\endgroup$ – BallBoy Nov 28 '18 at 0:49
  • $\begingroup$ @Y.Forman I've adjusted my answer accordingly. It seems our answers have converged. $\endgroup$ – Inactive - Objecting Extremism Nov 28 '18 at 0:54
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Let $f(x)=\frac1x$, $x_0=0$, then on the LHS we have $1$.

On the right hand side $\lim_{x \to x_0} f(x)$ is not defined.

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How about $x_0 = 0$, $f(x) = 1/x^2$?

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If both limits exist, the equality is true by the product rule of limits: $\lim_{x\to x_0} xf(x) = \lim_{x\to x_0} x \lim _{x\to x_0}f(x) = x_0 \lim _{x\to x_0}f(x)$

So the only counterexamples to equality would be cases one limit doesn't exist, e.g., $f(x)=\frac1x$ with $x_0=0$

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