Evaluating Limit for (x^3 + x + 2) / (x^4 -x +1) Tending Towards Infinity

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In summary, when evaluating the limit of (x^3 + x + 2) / (x^4 - x + 1) as x approaches infinity, the first step is to factor out x^4 from the numerator and denominator. This results in the limit of 1/x as x approaches infinity, which simplifies to 0. This is because the degree of the exponent in the denominator is larger than any in the numerator, causing the numerator to become insignificant at infinity. This can also be proven by factoring out x^4 from every term in the numerator and denominator and evaluating the limit. It is important to understand the concept behind this method rather than just memorizing the steps.
  • #1
fran1942
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Hello, I am just starting to learn limit evaluation techniques. I am unsure of the method used in this case.

(x^3 + x + 2) / (x^4 -x +1)
limit tending towards infinity.

I know the first step is 'x^3 / x^4', then '1/x = 0'
But I don't understand how this came about.

Can someone please clarify these steps.
Thanks kindly if possible.
 
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  • #2
You want a proof of the limit of 1/x is 0?
Do you know the definition of limit?

Another thing. "1/x = 0" is wrong its "1/x -> 0 as x -> infinity".
 
  • #3
Factor x4 out of every term in both the numerator and denominator, and then take the limit.
 
  • #4
normally teachers don't really look for the worked out solution.. they just expect you to know if it is 0 or infinity or just the coefficients. and that all depends on the greatest power of the denominator and numerator
 
  • #5
Basically you can just factor everything by x^4 like Mark44 says but then you'd have to factor it out of the constants too and you'll end up with a jumbled mess, which if you then evaluate the limit for you can prove is 0.

It's important for you to be able to conceptualize this and understand why it approaches 0.

The degree of the exponent in the bottom is larger than any in the top, therefore, infinity climbs faster at the bottom than it does at the top and at infinity the rate is infinitely greater thus the numerator eventually becomes insignificant.
 
  • #6
Well come.Please see these steps carefully.
(lim x...>∞(x3+x+1/x4-x+1)
taking x3 and x4 common 4m numerator & denominator respectively
=lim x...>∞[x3(1+1/x2+1/x3)/x4(1-1/x3+1/x4)]
x3 and x4 will cancel each other which comes 1/x
=lim x...>∞[1(1+1/x2+1/x3)/x(1-1/x3+1/x4)]
as by applying limit,1/x2 comes 1/(∞)2=1/∞=0 and so on,so
=1(1+0+0)/∞(1-0+0)
=1/∞
=0.
 

1. What is a limit?

A limit is a mathematical concept that describes the behavior of a function as its input approaches a certain value or "approaching infinity". It is often used to determine the behavior of a function at a specific point or to find the value of a function at a point where it is undefined.

2. How do I evaluate a limit?

To evaluate a limit, you can use several methods such as direct substitution, factoring, rationalization, and L'Hopital's rule. These methods involve manipulating the function algebraically to simplify it and then substituting the value the function is approaching into the simplified version to find the limit.

3. What is direct substitution?

Direct substitution is a method of evaluating a limit by substituting the value the function is approaching directly into the function. This method can only be used if the function is continuous at the given value.

4. What is L'Hopital's rule?

L'Hopital's rule is a method of evaluating limits of functions that involve indeterminate forms, such as 0/0 or infinity/infinity. It states that if the limit of a function f(x) divided by g(x) is an indeterminate form, then the limit of the derivative of f(x) divided by the derivative of g(x) will be the same.

5. Can limits be evaluated at all real numbers?

No, limits can only be evaluated at values where the function is defined. For example, if a function has a vertical asymptote at a certain value, the limit does not exist at that value. Additionally, limits can only be evaluated at values where the function is continuous.

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