# Example showing $\lim\limits_{x \to x_0} xf(x) \neq x_0\lim\limits_{x \to x_0} f(x)$

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?

• $\lim_{x \to 0} x\frac{1}{x}$, but if you insist that $\lim_{x \to x_o} f(x)$ needs to always be defined, then such an example doesn't exist – AlkaKadri Nov 28 '18 at 0:45
• 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. – Xander Henderson Nov 28 '18 at 0:45

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.

• Existence of the limit is enough for the product rule; continuity is not required. – BallBoy Nov 28 '18 at 0:49
• @Y.Forman I've adjusted my answer accordingly. It seems our answers have converged. – Inactive - Objecting Extremism Nov 28 '18 at 0:54

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.

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

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 $$x_0=0$$