Taking limits Watch

Big-Daddy
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I would like if someone could explain to me (or link me to) some decently rigorous procedures for taking the limits of functions in some simple cases.

I am mainly interested in taking the limits of x -> 0 and x -> infinity for the following cases:

(Let us define any g(x) or h(x) below as a polynomial, and the total function f(x) being the function I want to take the limit for

First priority is the case f(x)=g(x)/h(x). Then there is f(x)=eg(x)*h(x), followed by f(x)=ln(g(x))*h(x).

I have seen L'Hopital's Rule but that doesn't actually explain how to take the limit, only how to simplify the expression which you want to take the limit of.

Any help would be much appreciated.
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brianeverit
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(Original post by Big-Daddy)
I would like if someone could explain to me (or link me to) some decently rigorous procedures for taking the limits of functions in some simple cases.

I am mainly interested in taking the limits of x -> 0 and x -> infinity for the following cases:

(Let us define any g(x) or h(x) below as a polynomial, and the total function f(x) being the function I want to take the limit for

First priority is the case f(x)=g(x)/h(x). Then there is f(x)=eg(x)*h(x), followed by f(x)=ln(g(x))*h(x).

I have seen L'Hopital's Rule but that doesn't actually explain how to take the limit, only how to simplify the expression which you want to take the limit of.

Any help would be much appreciated.
If h(x) and g(x) are polynomials then for limit as x tends to zero, just put x=0 and see what you get. For example LImit\frac{x^2-x-2}{x^3-2x^2+3}=\frac{-2}{3}=-\frac{2}{3}

For x tends to infinity, If h(x) is of degree n and g(x) is of degree m then divide top and bottom by  x^k where k is the smaller of n and m, and use the fact that  \frac{1}{x^n} tends to zero for any positive integer n
For example LImit\frac{x^2-x-2}{x^3-2x^2+3}=\frac{1-\frac{1}{x}-\frac{2}{x^2}}{x-2+\frac{3}{x^2}} \mathrm\[\ which \ tends\ to\ }\frac{1}{x-2) so limit will be zero
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Big-Daddy
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(Original post by brianeverit)
If h(x) and g(x) are polynomials then for limit as x tends to zero, just put x=0 and see what you get. For example LImit\frac{x^2-x-2}{x^3-2x^2+3}=\frac{-2}{3}=-\frac{2}{3}
Thanks. So the limit will be the ratio of the terms in the numerator to the denominator which do not include x to any degree (except 0).

(Original post by brianeverit)
For x tends to infinity, If h(x) is of degree n and g(x) is of degree m then divide top and bottom by  x^k where k is the smaller of n and m, and use the fact that  \frac{1}{x^n} tends to zero for any positive integer n
For example LImit\frac{x^2-x-2}{x^3-2x^2+3}=\frac{1-\frac{1}{x}-\frac{2}{x^2}}{x-2+\frac{3}{x^2}} \mathrm\[\ which \ tends\ to\ }\frac{1}{x-2) so limit will be zero
I've had an idea: how about dividing by x^k where k is the larger of n and m, and then setting to infinity, so that any terms in the numerator or denominator each which do not contain x^k will go immediately to 0, but if you have an ax^k term in the numerator and bx^k in the denominator you will end up with a/b being the limit?

Meanwhile if the numerator has the larger of n and m, then the expression will go to infinity, whereas if the denominator has the larger of n and m it will go to 0. Am I correct here?

If so, could we move on - how do we evaluate something like the limits to 0 or infinity of f(x)=eg(x)*h(x)?
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lubus
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(Original post by Big-Daddy)
I would like if someone could explain to me (or link me to) some decently rigorous procedures for taking the limits of functions in some simple cases.

I am mainly interested in taking the limits of x -> 0 and x -> infinity for the following cases:

(Let us define any g(x) or h(x) below as a polynomial, and the total function f(x) being the function I want to take the limit for

First priority is the case f(x)=g(x)/h(x). Then there is f(x)=eg(x)*h(x), followed by f(x)=ln(g(x))*h(x).

I have seen L'Hopital's Rule but that doesn't actually explain how to take the limit, only how to simplify the expression which you want to take the limit of.

Any help would be much appreciated.
Firstly, lim(f(x)/g(x)) is limf(x)/limg(x)

Also, lim[f(x)g(x)] is limf(x)*limg(x)

This is known as the product and quotient rule for limits. The proof is an epsilon-delta proof.

if you have lim[e^g(x)] = e^limg(x)
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Hodor
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(Original post by Big-Daddy)
Thanks. So the limit will be the ratio of the terms in the numerator to the denominator which do not include x to any degree (except 0).



I've had an idea: how about dividing by x^k where k is the larger of n and m, and then setting to infinity, so that any terms in the numerator or denominator each which do not contain x^k will go immediately to 0, but if you have an ax^k term in the numerator and bx^k in the denominator you will end up with a/b being the limit?

Meanwhile if the numerator has the larger of n and m, then the expression will go to infinity, whereas if the denominator has the larger of n and m it will go to 0. Am I correct here?

If so, could we move on - how do we evaluate something like the limits to 0 or infinity of f(x)=eg(x)*h(x)?
For the limit as x goes to zero, note that polynomials and e^x are continuous functions, so you can just plug in x=0 to get the limit as h(0)e^g(0).

For the limit at infinity, perhaps use that g(x) and h(x) must go to either infinity or -infinity as x grows (depending in the sign of the leading term). This gives you four cases which are fairly straightforward to deal with.

I'm not really sure how much rigour you're after though.


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brianeverit
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(Original post by Big-Daddy)
Thanks. So the limit will be the ratio of the terms in the numerator to the denominator which do not include x to any degree (except 0).



I've had an idea: how about dividing by x^k where k is the larger of n and m, and then setting to infinity, so that any terms in the numerator or denominator each which do not contain x^k will go immediately to 0, but if you have an ax^k term in the numerator and bx^k in the denominator you will end up with a/b being the limit?

Yes

Meanwhile if the numerator has the larger of n and m, then the expression will go to infinity, whereas if the denominator has the larger of n and m it will go to 0. Am I correct here?


Yes
If so, could we move on - how do we evaluate something like the limits to 0 or infinity of f(x)=eg(x)*h(x)?
For e^{g(x)}h(x) If x tends to zero then this expression will tend to  ae^b where a and b are the constant terms in h(x) and g(x) respectively. It will pretty obviously tend to infinity as x tends to infinity.
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Big-Daddy
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(Original post by brianeverit)
For e^{g(x)}h(x) If x tends to zero then this expression will tend to  ae^b where a and b are the constant terms in h(x) and g(x) respectively.
Ok so once again simply substituting in x=0.

(Original post by brianeverit)
It will pretty obviously tend to infinity as x tends to infinity
I'm not so sure. What if the coefficients are competing - keep in mind that in the polynomials, e.g. ax^3 + bx^2 + cx + d, any coefficient a,b,c,d can be negative as well as positive...
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brianeverit
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(Original post by Big-Daddy)
Ok so once again simply substituting in x=0.



I'm not so sure. What if the coefficients are competing - keep in mind that in the polynomials, e.g. ax^3 + bx^2 + cx + d, any coefficient a,b,c,d can be negative as well as positive...
Well of course it could tend to negative infinity
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Big-Daddy
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(Original post by brianeverit)
Well of course it could tend to negative infinity
Or to 0, if the polynomial raised to the exponential overpowers the non-raised polynomial. And perhaps these are not the only possibilities. How should we go about evaluating it?
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brianeverit
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(Original post by Big-Daddy)
Or to 0, if the polynomial raised to the exponential overpowers the non-raised polynomial. And perhaps these are not the only possibilities. How should we go about evaluating it?
Of course, if g(x) tends to negative infinity..
Similarly with ln(g(x)*h(x) , ln(g(x) tends to positive infinity as x does, but negative infinity as x tends to zero, h(x) tends to + or - infinity as x does dependng on the sign attached top the highest power of x and to the term independent of x if x tends to zero.
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Big-Daddy
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(Original post by brianeverit)
Of course, if g(x) tends to negative infinity..
Similarly with ln(g(x)*h(x) , ln(g(x) tends to positive infinity as x does, but negative infinity as x tends to zero, h(x) tends to + or - infinity as x does dependng on the sign attached top the highest power of x and to the term independent of x if x tends to zero.
Ok so then let's summarize these rules for exp and ln functions:

e^(g(x))*(h(x))

If the coefficient on the highest-degree term in g(x) is positive, and on h(x) is positive, the limit to infinity will be infinity; if the former is positive but the latter negative, the limit to infinity will be -infinity; if the former is negative, then regardless of the sign on the latter, the limit will be 0.

The limit to 0 can be found by substituting x=0 into the function.

ln(g(x))*h(x)

If the coefficient on the highest-degree term in h(x) is positive, the limit to infinity will be infinity; if it is negative, the limit will be -infinity. i.e. g(x) does not matter here with regards to the infinite limit.

The x->0 limit is harder for this case. If there are terms without x, then ln(a)*b where a is the term without x in g(x) and b is the term without x in h(x) seems to be the infinite limit, achieved just by plugging in x=0. If there is no term a then perhaps we get ln(0)*b=-infinity as the limit to 0; if there is no term b then we get ln(a)*0=0. If there is neither a nor b then we get ln(0)*0 and I think this would come to 0.

Are these all right?
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brianeverit
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(Original post by Big-Daddy)
Ok so then let's summarize these rules for exp and ln functions:

e^(g(x))*(h(x))

If the coefficient on the highest-degree term in g(x) is positive, and on h(x) is positive, the limit to infinity will be infinity; if the former is positive but the latter negative, the limit to infinity will be -infinity; if the former is negative, then regardless of the sign on the latter, the limit will be 0.

The limit to 0 can be found by substituting x=0 into the function.

ln(g(x))*h(x)

If the coefficient on the highest-degree term in h(x) is positive, the limit to infinity will be infinity; if it is negative, the limit will be -infinity. i.e. g(x) does not matter here with regards to the infinite limit.

The x->0 limit is harder for this case. If there are terms without x, then ln(a)*b where a is the term without x in g(x) and b is the term without x in h(x) seems to be the infinite limit, achieved just by plugging in x=0. If there is no term a then perhaps we get ln(0)*b=-infinity as the limit to 0; if there is no term b then we get ln(a)*0=0. If there is neither a nor b then we get ln(0)*0 and I think this would come to 0.

Are these all right?
I think so.
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