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?!?!?!

No vote yet

3 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestAre you

sure\(\lim_{x\to0^+} g(x)=0\)?Log in to reply

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

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 \(\lim_{x\to n}g(x) \neq 0\)

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply