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# FP3 Limits, Is My Answer Sufficient ?

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Q:

Show that

I go:

As

Hence as the denominator tends towards then clearly

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

then obviously that is undetermined, but if we have

then can it be said that as

?

Am I wright in thinking we can't write

because the limiting value must be finite and so we write

or

as

Thnx
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Bump, anyone ?
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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.

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(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.

Right, so is my answer to the first part of the question correct and sufficient ?
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(Original post by member910132)
Q:

Show that

I go:

As

Hence as the denominator tends towards then clearly

Would that suffice as a valid answer ?
No. or are undefined.
Arranging F(x)

So if the form of the limit will be
like
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

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
In this set and

and f.e.
but if we have

then can it be said that as

?
AS I noted above if we have

and through the positive reals or
through the negative reals
are two cases.
For the first
for the second

that is the
limit is not exists.
Am I wright in thinking we can't write

because the limiting value must be finite and so we write

or

as

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

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