If I know that $\lim_{x \rightarrow 0} \frac{1+x}{x} = 1$, can I just "multiply through" by $x$ to conclude $\lim_{x \rightarrow 0} (1+x) = x$? I am worried about violating limit rules. I viewed this similar math.StackExchange question but the answers are not fully in agreement and rather complicated.
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4Your original limit isn't $1$, though. – Randall Jan 10 '23 at 20:13
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That limit is not $1$ but it does not exist. – Angelo Jan 10 '23 at 20:14
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You would have a contradiction if you multiplied by $x$: $\lim_{x \rightarrow 0} (1+x) = x \implies 1=0$. – Sebastiano Jan 10 '23 at 20:18
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1Never, ever claim $\lim_{x\to a}f(x)$ is anything other than a constant (or non-existent); it's not a non-constant function of $x$, which is just a dummy variable. What you can write is $f(x)\sim g(x)$ as $x\to a$. (As several people already noted, the example you have in mind is wrong.) – J.G. Jan 10 '23 at 22:56
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Suppose I were to start out a question with, "If I know that $1 = 0,$ can I ... ." How would you answer such a question? This is the kind of question you have asked here, namely, what happens if you "know" a fact but actually the fact is false? – David K Jan 14 '23 at 22:01
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You cannot do this. In a limiting expression like $\lim_{x \to 0} \frac{1+x}{x}$, the variable $x$ is a dummy variable (also called bound variable, or local variable in programming) that exists only within the limit expression $\frac{1+x}{x}$ itself. You cannot multiply by $x$ from the outside; $x$ doesn't exist there.
Ted
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