# If $\lim_{x \rightarrow 0}\frac{f(x)}{x^2} = 5$, then what is $\lim_{x \rightarrow 0}f(x)$?

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?

• 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. – InterstellarProbe May 8 at 16:38
• Do you have an example of another problem in which logic from the second method fails? – Smash May 8 at 16:45
• 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. – Théophile May 8 at 16:47

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 $$5x^2-\frac{x^2}n If $$x\to 0$$ we get $$f(x)\to 0$$.

$$\frac{\lim_{x \rightarrow 0}f(x)}{\lim_{x \rightarrow 0}x^2} = 5$$ 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 $$x^2$$.

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.

• You meant "denominator," not numerator. – Mark Viola May 8 at 16:41