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