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Two easy limit questions.

  1. Jun 4, 2005 #1
    I am having trouble with a couple of limit concepts in my class homework. For some reason my teacher says he wont help us or tell us the answer to the even questions, and he is not collecting them for homework either.:confused: He says your on your own with those questions, so i hope you can help me.

    First, my book says that a limit doesnt exist when the denominator tends to zero, but the numerator dosent. The answer i got was the numerator tends to 18 while the denominator tends to 0 and i thought this was infinity... Why is this not infinity? :confused:

    Second, lim as x->inf. of (ax/(sqr.root. of (x^2+ax) + x)) I am thinking that I need to factor out an x, but im not sure how to do this or if this is even what i should do....

    Thx for the help
  2. jcsd
  3. Jun 4, 2005 #2


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    Homework Helper

    Probably because

    [tex] \lim_{x \rightarrow 0^{-}} f(x) \neq \lim_{x \rightarrow 0^{+}} f(x) [/tex]

    A classic example is

    [tex] \lim_{x \rightarrow 0^{-}} \frac{1}{x} = - \infty [/tex]

    [tex] \lim_{x \rightarrow 0^{+}} \frac{1}{x} = + \infty [/tex]

    Therefore [itex] \lim_{x \rightarrow 0} \frac{1}{x} [/itex] does not exist.


    [tex] \lim_{x \rightarrow \infty} \frac{ax}{\sqrt{x^2+ax + x}} [/tex]

    rewrite it as

    [tex] \lim_{x \rightarrow \infty} \frac{ax}{x \sqrt{1 + \frac{a}{x} + \frac{1}{x}}} [/tex]
  4. Jun 4, 2005 #3
    A number infinitely small (as close to zero without being zero) would give you infinity. However, this limit can specifically reach zero. But a number divided by zero doesn't have a solution. The limit of any fraction where the denominator = 0 (watch out for cases when you can factor out something to get rid of this) is not going to exist.

    For the second part. Yes, you do need to factor out an x...but a little more. Focus on what is in the square root. You have [tex]x^2 + ax[/tex]. Just because the smallest degree of x is 1, it doesn't mean you can't factor out more. Try factoring out an [tex]x^2[/tex] and continue on.
  5. Jun 4, 2005 #4
    Thx, that helps... but perhaps i should be more specific about the first part.

    First it is given that lim x->+inf. of f(x)= 3 and of g(x)=-5

    then i find the limit x->+inf. of 6f(x)/(5f(x)+3g(x))

    what i get is 18/0 as x ->+inf. and i think this means that as the denominator gets closer to 0, the value gets larger.

    sorry this is a dumb question
  6. Jun 4, 2005 #5
    about the second part - acutally the last x isnt in the radical.

    I get it now I was confused as how the x factors out of the radical...
  7. Jun 4, 2005 #6


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    Ok about the first part again, You cannot compute the limit with limit laws, because the limit of the denominator tends to 0, and the limit laws of the quotient require the limit of the denominator does not tend to 0, so it's simply said the limit does not exist.

    The problem with the limit laws is that it assumes the limits of f(x) and g(x) exist and are finite.
    Last edited: Jun 4, 2005
  8. Jun 4, 2005 #7


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    "Infinity" is not a number. Saying that "the limit is not infinity" is just saying that the limit does not exist in a particular way. If the denominator of a fraction goes to 0 and the numerator does not, "the limit is infininity" and "the limit does not exist" may both be true.
  9. Jun 5, 2005 #8
    Thank you.
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