# How should I interpret the product of limits $\lim_{x\rightarrow1}(x-1)\lim_{x\rightarrow1}\frac{1}{x-1}$

$$\lim_{x\rightarrow1}(x-1)\lim_{x\rightarrow1}\frac{1}{x-1}$$

I know that the first limit is just $$0$$

However, I am confused since the second limit diverges.

Should the answer be $$0$$ (since the first limit is $$0$$), or should it be undefined? (since the second limit does not exist)

WolframAlpha says it is undefined.

So is it $$0$$, or undefined?

• I believe this is meant to be a trick question. By the limit laws, the product of the limits is equal to the limit of the product (so long as all limits exist). So, the problem is trying to confuse you into trying $$\lim_{x \to 1} \dfrac{x-1}{x-1} = 1$$ which is wrong. WolframAlpha is correct. $$\dfrac{0}{0}$$ is undefined. – InterstellarProbe Jan 24 '20 at 14:42
• @InterstellarProbe How did the expression become $\frac{0}{0}$? – Pizzaroot Jan 24 '20 at 14:48
• the first limit is approaching $0$ while the second is approaching $\dfrac{1}{0}$. Their product is an indeterminate form. This is not a formal answer, which is why I am only putting it in the comments, but that is the intuition behind it. – InterstellarProbe Jan 24 '20 at 14:48

The expression you wrote is undefined. If you were trying to evaluate: $$\lim_{x\to 1} \frac{x-1}{x-1}$$ you cannot split into product of limits. $$\lim_{x\to 1} \frac{x-1}{x-1}\neq \ \lim_{x\to 1} (x-1)\lim_{x\to 1} \frac{1}{x-1}$$
Basic analysis theorems tell you: if two functions $$f$$ and $$g$$ admit limits and they are finite, then the limit of the product is the product of the limit.
What is your expression? The first factor is $$0\in\mathbb R$$, but $$∞\notin \mathbb R$$. The product $$0\cdot ∞$$ is not well defined, not even inside $$\tilde{\mathbb{R}}$$, the extended real line, which is a fancy thing, but not a ring where the operation "product" is well defined.