Limits Involving Infinity

Key Questions

  • How do I find the limit as x approaches infinity of the square root function?

    Answer:

    lim_(x->oo)sqrt(x) = oo

    Explanation:

    Intuitively, as there is no bound to how large we can make sqrt(x) by increasing x, we expect that the limit as x->oo of sqrt(x) would be oo. Indeed, if there were such a bound, say x_0, then we would arrive at a contradiction, as sqrt(x_0^2+1) > sqrt(x_0^2)= x_0.

    We can, however, approach the problem in a more rigorous manner.

    We say that the limit as x->oo of a function f(x) is oo (alternately f(x)->oo as x->oo), denoted lim_(x->oo)f(x)=oo, if, for every integer N>0 there exists an integer M>0 such that x>M implies f(x)>N.

    Less formally, that means that for any real value, f(x) will be greater than that value for large enough x.

    Our claim is that lim_(x->oo)sqrt(x) = oo. Let's prove it using the above definition.

    Take any integer N>0, and let M=N^2. Then, for any x>M, we have

    sqrt(x) >sqrt(M) = sqrt(N^2) = N

    We have shown that for any integer N>0 there exists an integer M>0 such that x>M implies sqrt(x) > N, thereby proving that lim_(x->oo)sqrt(x) = oo.

    The above method actually can be used to show that x^k->oo as x->oo for any k>0. If we start with an arbitrary N>0 and let M=N^(1/k), then for x>M we have x^k > M^k = (N^(1/k))^k=N. As sqrt(x) = x^(1/2), the above is just a special case of this.

    sente · 4 · May 27 2016
  • What is the limit as x approaches infinity of x?

    Answer:

    Lim_(x->oo)x=oo

    Explanation:

    Break the problem down into words: "What happens to a function, x, as we continue increasing x without bound?"

    x would also increase without bound, or go to oo.

    Graphically, this tells us that as we continue heading right on the x-axis (increasing values of x, going to oo) our function, which is just a line in this case, keeps heading upwards (increasing) with no restrictions.

    graph{y=x [-10, 10, -5, 5]}

    VNVDVI · 1 · Mar 2 2018
  • How do I find the limit as x approaches negative infinity of the square root function?

    The answer is undefined.

    lim_(x->-oo)sqrt(x)=undefined

    The reason for this is that the domain of the square root function is x>=0. The limit is undefined if the limit is not being evaluated in the domain.

    Amory W. · · Aug 29 2014
  • How do I find the limit as x approaches negative infinity of a polynomial?

    Answer:

    I usually do this informally.

    Explanation:

    I ask myself what kinds of numbers do I get if I put in more and more negative numbers for x. (Often described by saying "bigger and bigger negative numbers".) ("Big" means "far from zero".)

    Examples:

    Example 1
    f(x) = 3x^4-7x^3+2x+72

    For very very big numbers, the only term that matters is the largest power term: 3x^4. As I put in bigger and bigger negatives, do I get bigger and bigger positives or negatives?

    x^4 will always be positive and when I multiply by 3 the answer will still be positive, so I get bigger and bigger positive numbers.

    lim_(xrarr-oo)f(x) = oo

    Example 2
    g(x) = 5x^7+43x^4+2x^3-5x+21

    For very very big numbers, the only term that matters is the largest power term: 5x^7. As I put in bigger and bigger negatives, do I get bigger and bigger positives or negatives?

    x^7 will be negative for negative x's and when I multiply by 5 the answer will still be negative, so I get bigger and bigger negative numbers.

    lim_(xrarr-oo)g(x) = -oo

    Example 3 (last)
    h(x) = -8x^6+7x-3

    For very very big numbers, the only term that matters is the largest power term: -8x^6. As I put in bigger and bigger negatives, do I get bigger and bigger positives or negatives?

    x^6 will be positive for all x's and when I multiply by -8 the answer will become negative, so I get bigger and bigger negative numbers.

    lim_(xrarr-oo)h(x) = -oo

    Jim H · 2 · Jul 15 2015

Questions

Limits