# Limit of Equality

Hey, I have stumbled upon something quite interesting. If $$\lim_{x\to n} \frac{f\left(x\right)}{g\left(x\right)} = 1$$, that means that $$\lim_{x\to n} f\left(x\right) = \lim_{x\to n} g\left(x\right)$$. This assumption may be wrong. But, for $$f\left(x\right) = \ln x$$ and $$g\left(x\right) = \left(x^2+1\right)\ln x$$, $$\lim_{x\to 0^+} f\left(x\right) = -\infty$$ and $$\lim_{x\to 0^+} g\left(x\right) = 0$$. But, even stranger is that $$\lim_{x\to 0^+} \frac{f\left(x\right)}{g\left(x\right)} = 1$$.

WHAT IS GOING ON?!?!?!

Note by Jimmi Simpson
5 years, 7 months ago

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- 5 years, 7 months ago

Thats a good observation. Apart from using Wolfram, we know that exponential growth dominates any polynomial growth, which is why we have $$\lim_{x \rightarrow 0^+} g(x) = - \infty$$.

Staff - 5 years, 7 months ago

Oh, I see. I had checked by graphing only to realise that I had entered the wrong function.

- 5 years, 7 months ago

I think that assumption is wrong because $$\lim_{x\to n}\frac{f(x)}{g(x)}=\frac{\lim_{x\to n}f(x)}{\lim_{x\to n}g(x)}$$ only when $$\lim_{x\to n}g(x) \neq 0$$

- 5 years, 7 months ago

Yes, always make sure you that the conditions of the theorem are satisfied before you apply it.

Staff - 5 years, 7 months ago

Okay, then flip it. Either way the limit is 1.

- 5 years, 7 months ago

@Cody Look up the actual conditions of the theorem. Flipping it doesn't help since neither $$f(x)$$ nor $$g(x)$$ satisfy the conditions.

Staff - 5 years, 7 months ago