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#1 2007-06-02 01:25:40

nox_populi
Member
Registered: 2007-06-02
Posts: 2

Calculus: some problem with limits...

Hi,
I have this little problem. It has been bugging me a while. I still at high school so its nothing advanced. I tried l'Hopital rule, log, differentiation, etc. and nothing of it worked. Here it is:

Limit[(2^{x + 3} + 4)/(2^{x - 1} + 1), x -> Infinity]

The problem is that it has this wierd exponent. Anyone has any ideas? smile
Cheers,
Dan

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#2 2007-06-02 03:11:03

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Calculus: some problem with limits...

As x approaches ∞, the power of 2 terms get very large, and so the constant terms can be neglected. After that it's just using power laws and working out 2^4.

For future reference, L'Hopital is used when you need to find the limit of a fraction that would be 0/0 if evaluated (usually as x -> 0). If the fraction is ∞/∞ then you have to use other tricks.


Why did the vector cross the road?
It wanted to be normal.

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#3 2007-06-02 05:01:33

Identity
Member
Registered: 2007-04-18
Posts: 934

Re: Calculus: some problem with limits...

So there is no single algorithm for simplifying limits, but a whole bunch of different methods? Are there any limits that exist but cannot be solved?

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#4 2007-06-02 12:56:09

nox_populi
Member
Registered: 2007-06-02
Posts: 2

Re: Calculus: some problem with limits...

Thanks mathsyperson smile ...just btw what are power laws? Maybe I know but perhaps the translation differs in my language from english so I dont know what it should be exactly.

Can I do it also this way(?):

according to the rule that [x^a * x^r = x^{a+r}] so that our example will be like [2^x * 2^3]/[2^x * 2^-1] and that this way the [2^x] will cancel each other out and we will get [2^3]/[2^-1] which equals eventually [16] ...

Last edited by nox_populi (2007-06-02 12:57:56)

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#5 2007-06-02 14:08:06

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: Calculus: some problem with limits...

Standard proof:

When f(x)->a and g(x)->b and b≠0
lim[f(x)/g(x)]=a/b

Last edited by George,Y (2007-06-02 14:11:41)


X'(y-Xβ)=0

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#6 2007-06-02 19:33:22

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Calculus: some problem with limits...

Ah, true enough. You're technically meant to neglect things because they're infinitessimal, rather than because they're constant and infinite terms are present.

And the power laws I was talking about are pretty much exactly what you just did in your post. smile


Why did the vector cross the road?
It wanted to be normal.

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#7 2007-06-02 20:03:32

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: Calculus: some problem with limits...

I see, the power law is something like a shortcut.


X'(y-Xβ)=0

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