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I know the answer to the above question, but I have a question on some of the reasoning.

The way I know how to solve it is $$\lim_{x \rightarrow 0}f(x) = \lim_{x \rightarrow 0}\left(f(x)\cdot \frac{x^2}{x^2}\right) = \lim_{x \rightarrow 0}\left(\frac{f(x)}{x^2}\cdot x^2\right) = \left(\lim_{x \rightarrow 0}\frac{f(x)}{x^2}\right)\left(\lim_{x \rightarrow 0}x^2\right) = 5\cdot0 = 0.$$

I saw another solution elsewhere that gets the right answer, but I am unsure if the steps are actually correct. \begin{align*} &\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = 5 \\ \Longrightarrow &\frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2} = 5 \\ \Longrightarrow &\lim_{x \rightarrow 0}f(x) = 5\cdot \lim_{x \rightarrow 0}x^2\\ \Longrightarrow &\lim_{x \rightarrow 0}f(x) = 5\cdot 0 = 0. \end{align*}

My issue is with that first step. I know that $\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow a}f(x)}{\lim_{x \rightarrow a}g(x)}$, but only when $\lim_{x \rightarrow a}g(x) \neq 0$. Since $\lim_{x \rightarrow 0}x^2 = 0$, wouldn't this invalidate the above work? However, it still got the same answer, so my real question is why did it work and when will it work in general?

EDIT: Does anyone have a nice example for when the logic in the second method doesn't work?

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  • $\begingroup$ You are correct that the solution you saw elsewhere is invalid. It is invalid for the exact reason you think it is. Using false logic to get a correct answer is still false logic. Applied to another problem, it might fail. $\endgroup$ – InterstellarProbe May 8 '19 at 16:38
  • $\begingroup$ Do you have an example of another problem in which logic from the second method fails? $\endgroup$ – user525033 May 8 '19 at 16:45
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    $\begingroup$ Simply put, you can't divide by zero. If you want examples, there are many "proofs" of $0=1$ where the trick is to divide by zero and hope no one notices. $\endgroup$ – Théophile May 8 '19 at 16:47
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If $\lim\limits_{x\to0}\dfrac{f(x)}{x^2}=5$, then for every $n\in\Bbb N$ there is some $x\neq 0$ such that $$5-\frac1n<\frac{f(x)}{x^2}<5+\frac1n$$ and hence $$5x^2-\frac{x^2}n<f(x)<5x^2+\frac{x^2}n$$ If $x\to 0$ we get $f(x)\to 0$.

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$$\frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2} = 5$$ doesn't work, the LHS is not defined.


But for the limit of $\dfrac{f(x)}{x^2}$ to exist, $\lim_{x\to0}f(x)$ must be zero, which is also the limit of $x^2$.

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You can not do this since the denominator on the right is $0$.

$$\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = \frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2}$$

But first aproach is perfect.

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    $\begingroup$ You meant "denominator," not numerator. $\endgroup$ – Mark Viola May 8 '19 at 16:41

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