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My calculus book, Differential and Integral Calculus by Piskunov, states the following theorem: theorem

This theorem is used in evaluating the following limit: $$\lim_{x \to 1}\dfrac{x}{1-x}$$ We first take the limit of the reciprocal of $\frac{x}{1-x}$ $$\lim_{x\to 1}\dfrac{1-x}{x}=0$$ Then, by the above theorem, $\lim_{x \to 1}\frac{x}{1-x} = \infty$.

$\frac{1-x}{x}$ approaches $0$ as $x \to 1$. But clearly, if we substitute $1$ for $x$, $\frac{1-x}{x}$ also becomes zero. How is the theorem applicable then?

I'm sure I'm not understanding the difference between approaching and becoming $0$ even after reading several related answer. Can you explain it in the context of this example?

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  • $\begingroup$ If the expression is valid at the given point, you can (usually) just insert the number to get the limit. But often the expression is invalid at the given point. Sometimes the limit exists nevertheless and sometimes (like here) not. $\endgroup$
    – Peter
    Apr 20, 2021 at 17:36
  • $\begingroup$ I think they mean that you don't hit a different divide by zero point as you approach your limit. Kind of like, "take these steps to approach that hole in the function, but be careful not to fall in a different hole on your way." $\endgroup$ Apr 20, 2021 at 17:37

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The point is that $\alpha(x) \neq 0$. The statement is that $\alpha(x) \to 0$, but $\alpha(x) \neq 0$ for all $x$ in a neighborhood of $a$ (not counting $a$ itself, which doesn't affect the limit).

To see why this would be a problem, suppose that $\alpha(x) = 0$ in a tiny ball $(a - \epsilon, a + \epsilon)$. Then it's not true that $\lim_{x \to a} 1/\alpha(x) = \infty$, as in fact the limit doesn't exist at all.

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Since the goal is to study the limit$$\lim_{x\to1}\frac{1-x}x,$$the value that $\frac{1-x}x$ takes when $x=1$ is irrelevant. For instance, if $f(x)=x$ and$$g(x)=\begin{cases}1&\text{ if }x=0\\x&\text{ otherwise,}\end{cases}$$then $f(x)=g(x)$ when $x\ne0$. So, since $\lim_{x\to0}f(x)=0$, you also have $\lim_{x\to0}g(x)=0$, in spite of the fact that $g(0)=1$.

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