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FP3 Limits, Is My Answer Sufficient ?

Q:

Show that F(x)→−∞ as x→0 \displaystyle F(x) \to -\infty \ \text {as} \ x\to 0

F(x)=1x3(1+ln⁡x) F(x) = \displaystyle \dfrac{1}{x^3(1+\ln x)}

I go:

1x3(1+ln⁡x)=1x3+x3ln⁡x) \displaystyle \dfrac{1}{x^3(1+\ln x)} = \dfrac{1}{x^3+x^3\ln x)}

As x→0 then x3→0 and x3ln⁡x→0− \displaystyle x\to 0 \ \text {then} \ x^3\to 0 \ \text {and} \ x^3\ln x \to 0-

Hence as the denominator tends towards 0− \displaystyle 0- then clearly F(x)→−∞ \displaystyle F(x)\to -\infty

Would that suffice as a valid answer ?

Secondly, and I would highly appreciate it if people could also answer this so that is completes my knowledge in the topic, if we have

10 \displaystyle \dfrac{1}{0} then obviously that is undetermined, but if we have

Unparseable latex formula:

\displaystyle \dfrac{1}{x} \and\ x\to 0

then can it be said that as

x→0 \displaystyle x\to 0

1x→∞ \displaystyle \dfrac {1}{x} \to \infty ?

Am I wright in thinking we can't write

lim⁡x→01x=∞ \displaystyle\lim_{x\to 0} \dfrac {1}{x} = \infty because the limiting value must be finite and so we write

lim⁡x→01x→∞ \displaystyle\lim_{x\to 0} \dfrac {1}{x} \to \infty or

1x→∞ \displaystyle\dfrac {1}{x} \to \infty as x→0 \displaystyle x\to 0


Thnx:smile:
(edited 11 years ago)
Reply 1
Avatar for member910132
member910132
OP
Bump, anyone ?
Reply 2
Avatar for Jodin
Jodin
2
Hi, firstly, it's fine to write that lim x ->0 1/x = infinity, or rather it would be if it were true (consider a limit approaching from the negative numbers (1/-1, 1/-0.1, 1/-0.01, ...)

You're saved in this limit because you're restricted to x>0 'cause of the lnx.

Edit: The edits are because I massively misread your post, sorry about that.
(edited 11 years ago)
Reply 3
Avatar for member910132
member910132
OP
Original post by Jodin
Hi, firstly, it's fine to write that lim x ->0 1/x = infinity, or rather it would be if it were true (consider a limit approaching from the negative numbers (1/-1, 1/-0.1, 1/-0.01, ...)

You're saved in this limit because you're restricted to x>0 'cause of the lnx.

Edit: The edits are because I massively misread your post, sorry about that.


Right, so is my answer to the first part of the question correct and sufficient ?
Reply 4
Avatar for ztibor
ztibor
10
Original post by member910132
Q:

Show that F(x)→−∞ as x→0 \displaystyle F(x) \to -\infty \ \text {as} \ x\to 0

F(x)=1x3(1+ln⁡x) F(x) = \displaystyle \dfrac{1}{x^3(1+\ln x)}

I go:

1x3(1+ln⁡x)=1x3+x3ln⁡x) \displaystyle \dfrac{1}{x^3(1+\ln x)} = \dfrac{1}{x^3+x^3\ln x)}

As x→0 then x3→0 and x3ln⁡x→0− \displaystyle x\to 0 \ \text {then} \ x^3\to 0 \ \text {and} \ x^3\ln x \to 0-

Hence as the denominator tends towards 0− \displaystyle 0- then clearly F(x)→−∞ \displaystyle F(x)\to -\infty

Would that suffice as a valid answer ?

No. 0⋅∞\displaystyle 0\cdot \infty or 0⋅(−∞)\displaystyle 0\cdot (-\infty) are undefined.
Arranging F(x)
F(x)=1x31+lnx\displaystyle F(x)=\frac{\frac{1}{x^3}}{1+lnx}
So if x→0+\displaystyle x\rightarrow 0^{+} the form of the limit will be
like
Unparseable latex formula:

\diaplaystyle \frac{\infty}{\infty}


Use the L'Hospital rule.


Secondly, and I would highly appreciate it if people could also answer this so that is completes my knowledge in the topic, if we have

10 \displaystyle \dfrac{1}{0} then obviously that is undetermined,

..depending on the number set you are using.
For calculating limit we can use an extension of Reals including infinte ordinals
that is R∪[−∞,∞]\displaystyle \mathbb{R} \cup [-\infty, \infty]
In this set 10−=−∞\displaystyle \frac{1}{0^-}=-\infty and
10+=∞\displaystyle \frac{1}{0^+}=\infty
and f.e. c±∞=0\frac{c}{\pm \infty}=0

but if we have

Unparseable latex formula:

\displaystyle \dfrac{1}{x} \and\ x\to 0

then can it be said that as

x→0 \displaystyle x\to 0

1x→∞ \displaystyle \dfrac {1}{x} \to \infty ?

AS I noted above if we have
1x\displaystyle \frac{1}{x}
and
Unparseable latex formula:

\isplaystyle x \to 0^+

through the positive reals or
x→0−\displaystyle x \to 0^- through the negative reals
are two cases.
For the first 1x→+∞\displaystyle \frac{1}{x} \to +\infty
for the second 1x→−∞\displaystyle \frac{1}{x} \to -\infty

that is the limx→01x\displaystyle lim_{x \to 0} \frac{1}{x}
limit is not exists.

Am I wright in thinking we can't write

lim⁡x→01x=∞ \displaystyle\lim_{x\to 0} \dfrac {1}{x} = \infty because the limiting value must be finite and so we write

lim⁡x→01x→∞ \displaystyle\lim_{x\to 0} \dfrac {1}{x} \to \infty or

1x→∞ \displaystyle\dfrac {1}{x} \to \infty as x→0 \displaystyle x\to 0


Thnx:smile:


No. This limit is not exists in this form (in your Q x>0 because of lnx)
(edited 11 years ago)

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