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?!?!?!

Easy Math Editor

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Comments

Are you sure $\lim_{x\to0^+} g(x)=0$ ?

o b - 8 years, 4 months ago

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

Calvin Lin Staff - 8 years, 4 months ago

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

Jimmi Simpson - 8 years, 4 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 $\lim_{x\to n}g(x) \neq 0$

Alan Zhang - 8 years, 4 months ago

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

Calvin Lin Staff - 8 years, 4 months ago

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

Jimmi Simpson - 8 years, 4 months ago

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

Calvin Lin Staff - 8 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$