I need help on a limit question involving infinity.?

[stefany]

lim of x tends to +infinity, (2-3x+4x^2)/(x^3+7)

I don't even know what to do when you plug it in to find out if it's undefined.

[cidyah]

You can tell by inspection that the limit is 0, since the exponent of the denominator is higher than that of the numerator.

L'Hopital's Rule:
Differentiate the numerator and the denominator.
lim x-> inf (-3+8x) / (3x^2)
lim x->inf 8 / 6x
as x goes to infinity, 8 / 6x goes to 0.

[agent177]

The general trick for this type of limit (one polynomial above another) is to divide top and bottom by the highest degree of x.

So in this example, divide top and bottom by x^3 to get:

(2-3x+4x^2)/(x^3+7) = (2/(x^3) - 3/(x^2) + 4/x)/(1 + 7/(x^3))

All terms in the top tend to 0, the bottom tends to 1, so the whole thing tends to 0/1 = 0.

[Dot]

There are a few rules that you should memorize for limits as rational polynomials go to infinity.

There are three basic rules for evaluating limits at infinity for a rational function f(x) = p(x)/q(x):
If the degree of p is greater than the degree of q, then the limit is positive or negative infinity depending on the signs of the leading coefficients;
If the degree of p and q are equal, the limit is the leading coefficient of p divided by the leading coefficient of q;
If the degree of p is less than the degree of q, the limit is 0.
If the limit at infinity exists, it represents a horizontal asymptote at y = L. Polynomials do not have horizontal asymptotes; they may occur with rational functions.

I'm going to add one more... it doesn't apply in your case, but it may apply to some of your homework.

If the degree of p is greater than q by EXACTLY 1 then a slant asymptote may exist. The equation of which is found by dividing p by q. The equation of the slant asymptote is given by y = q(x), where q is the quotient of the division.


In this particular case, the degree of p is less than that of q so the limit is zero, or there is a horizontal asymptote at y = 0

[CliffordR]

The Easy way: look at the degree of the polynomials. the highest degree of the numerator is 2, and it is 3 for the denominator. the limit as x goes to infinity is 0

rules for rational equations and limits as x -> infinity:

1. if the degree of the numerator is less then the degree of the denominator, the limit as x approaches infinity is 0

2. if the degrees are the same, divide the leading coefficient of the num. by the leading coeff. of the denom. and that will be the limit as x approaches infinity.

3. if the numerator's degree is higher than the denominator's, the limit as x goes to infinity is infinity

[hevans1944]

Or, just divide every term in the numerator and denominator by x^3 and inspect each term as x-> +infinity. In the numerator each term goes to zero; in the denominator the first term goes to one and the second to zero. So the limit is 0/1 or 0.

[DavidG]

hey mate,

the method provided by cidyah is certainly valid, however I found limits involving this type of expression can be attacked in a simpler manner

here you have

lim( x --> inf ) (2-3x+4x^2)/(x^3+7)
In this case the higher order is 3 (x^3), so divide top and bottom by x^3
i.e.
lim( x --> inf ) (1/x^3)(2-3x+4x^2)/((1/x^3)(x^3+7))
= lim( x --> inf ) (2/x^3-3/x^2+4/x^3)/(1+7/x^3)
= ( 0 + 0 + 0)/( 1 + 0) = 0/1 = 0

As stated before, the previous method posted is 100% valid, however when it involves polynomials it is sometimes 'easier' to divide the top and bottom by x to the power of the higher order.

Hope this helps,

David

[EricH]

Given :

lim (1/x) = 0
x --> ?

How that helps...

First multiply f(x) by (1/x^3)/(1/x^3).

Resulting in (4/x - 3/x^2 + 2/x^3) / (1 + 7/x^3)

Now you can take the limit by plugging in ?.

(4/? - 3/? + 2/?) / (1 + 7/?)

Using what we know this is equal to

(0 - 0 + 0) / (1 + 0) = 0/1 = 0

The trick for limits to ? like this one is to multiply the top and bottom by the reciprocal of highest power in the denominator. In this case the highest power was x^3.

[DAVE]

Take the first derivative of the top, and divide it by the first derivative of the bottom. If this is still undefined, repeat until it isn't. This is called "L'Hôpital's Rule".

Good luck!

[Dot]

There are a few rules that you should memorize for limits as rational polynomials go to infinity.

There are three basic rules for evaluating limits at infinity for a rational function f(x) = p(x)/q(x):
If the degree of p is greater than the degree of q, then the limit is positive or negative infinity depending on the signs of the leading coefficients;
If the degree of p and q are equal, the limit is the leading coefficient of p divided by the leading coefficient of q;
If the degree of p is less than the degree of q, the limit is 0.
If the limit at infinity exists, it represents a horizontal asymptote at y = L. Polynomials do not have horizontal asymptotes; they may occur with rational functions.

I'm going to add one more... it doesn't apply in your case, but it may apply to some of your homework.

If the degree of p is greater than q by EXACTLY 1 then a slant asymptote may exist. The equation of which is found by dividing p by q. The equation of the slant asymptote is given by y = q(x), where q is the quotient of the division.


In this particular case, the degree of p is less than that of q so the limit is zero, or there is a horizontal asymptote at y = 0