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?!?!?!

## Comments

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sure\(\lim_{x\to0^+} g(x)=0\)? – O B · 3 years, 11 months agoLog in to reply

– Calvin Lin Staff · 3 years, 11 months ago

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

– Jimmi Simpson · 3 years, 11 months ago

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

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 \(\lim_{x\to n}g(x) \neq 0\) – Alan Zhang · 3 years, 11 months ago

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– Calvin Lin Staff · 3 years, 11 months ago

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

– Jimmi Simpson · 3 years, 11 months ago

Okay, then flip it. Either way the limit is 1.Log in to reply

– Calvin Lin Staff · 3 years, 11 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.Log in to reply