Year 12s: what will be the first way you research your future options? >>

Mega A Level Maths Thread - Mark V

Scroll to see replies

Reply 280

Zacken

Original post by RoseGatz

Got a question I'm a bit stuck on, care to help @Zacken or anyone else for that matter? It's more pure stuff :flutter:

The function

Unparseable latex formula:

\displaystyle \begin{equation*}f\end{equation*}

is defined by

Unparseable latex formula:

\displaystyle \begin{equation*}f(x)= \frac{1}{x^3(1+Inx)}\end{equation*}

where

Unparseable latex formula:

\displaystyle \begin{equation*}x>0\end{equation*}

Show that

Unparseable latex formula:

\displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*}

Unparseable latex formula:

\displaystyle \begin{equation*}x \rightarrow{0}\end{equation*}

I've managed to show that

Unparseable latex formula:

\displaystyle \begin{equation*}f(x) \rightarrow\infty\end{equation*}

instead of showing

Unparseable latex formula:

\displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*}

, got no idea where the minus is coming from.

Impressive

\LaTeX

\displaystyle \lim_{x \to 0} \ln x = -\infty \Rightarrow \lim_{x\to 0} (1 + \ln x) = -\infty \Rightarrow \lim_{x \to 0} x^3(1 + \ln x) = 0

but in the negative sense, if you get me?

Look at the graph of

\ln x

for

0 < x < 1

to see why

\ln x

diverges to

-\infty

.

Although to be fair - if you have a limit that equals either plus or minus infinity, it's pretty much the same thing in that the limit does not exist.

Reply 281

RoseGatz

Original post by Zacken

Impressive

\LaTeX

\displaystyle \lim_{x \to 0} \ln x = -\infty \Rightarrow \lim_{x\to 0} (1 + \ln x) = -\infty \Rightarrow \lim_{x \to 0} x^3(1 + \ln x) = 0

but in the negative sense, if you get me?

Look at the graph of

\ln x

for

0 < x < 1

to see why

\ln x

diverges to

-\infty

Haha I know right! Took me a while to get the hang of it.

We went through this is class earlier today, we knew it was obviously

-\infty

considering the graph but is this just an assumption we make? Is there no way to explicitly get

Unparseable latex formula:

\displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*}

or do we just go off assumption because of the graph?

Original post by Zacken

Although to be fair - if you have a limit that equals either plus or minus infinity, it's pretty much the same thing in that the limit does not exist.

Reply 282

Zacken

Original post by RoseGatz

Haha I know right! Took me a while to get the hang of it.

We went through this is class earlier today, we knew it was obviously

-\infty

considering the graph but is this just an assumption we make? Is there no way to explicitly get

Unparseable latex formula:

\displaystyle \begin{equation*}f(x) \rightarrow{-\infty}\end{equation*}

or do we just go off assumption because of the graph?

Saying that

\lim_{x \to a} f(x) = \infty

is a shorthand way of saying that the limit doesn't exist, i.e: that it diverges. Having anything

=\infty

nearly almost always never makes sense. So fussing about whether it's plus or minus infinity is something that we don't really tend to do because in either case - we're still just saying that the limit doesn't exist.

If you want some justification for why

\lim_{x \to 0} \ln x = -\infty

, you could try looking at

\lim_{x \to -1} \ln (1 + x) = \lim_{x \to -1} (x - x^2 /2 + x^3/3 - \cdots )

, might work - I haven't thought about it properly though, I'm rushing this out.

Most of the things you do with limits is based on epsilon-delta proofs, that is, the formal condition for limits and once you start working with infinity in there, that falls apart because you can't have something converge to infinity - so at this level, you're reduced to looking at graphs and making assumptions instead of working with things like you would in a real analysis course.

This: http://math.stackexchange.com/questions/127689/why-does-a-limit-at-infinity-not-exist may shed some light on what I meant.

Reply 283

drandy76

Isn't the x^3 responsible for its divergence to -infinity, since as x approaches 0 ln(x) becomes large and negative therefore making the fraction small and negative?

Reply 284

Zacken

Original post by drandy76

Isn't the x^3 responsible for its divergence to -infinity, since as x approaches 0 ln(x) becomes large and negative therefore making the fraction small and negative?