Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Limits in general

  1. Mar 11, 2006 #1
    So monday I have my math test, it's gonna be about limits, derivatives and combination & permutation ...
    I have some question;
    if you get an indeterminate form of +[tex]\infty - \infty[/tex], is there any rule to ease everything for you. like how you can use the Hopital ruel for 0/0 ? anw what do you do in general when you get this kind of indeterminate form?
    if they ask you to see if the function is derivable in [tex]\x_{0}[/tex], when do you study the right and left derivation ?
    that s it for now!
    hope to hear from you soon. Thanks
  2. jcsd
  3. Mar 11, 2006 #2


    User Avatar
    Homework Helper

    Uhmm, can you be little bit clearer? Like giving us an example, or something along those lines...
    If you were ask to find the limit of some radicals, then it's common to multiply and divide the whole thing by its conjugate expression.
    [tex]\lim_{x \rightarrow +\infty} \sqrt{x ^ 2 + x} - x[/tex]
    Now, it's the Indeterminate form [tex]+ \infty - \infty[/tex], right?
    Multiply the whole expression with [tex]\frac{\sqrt{x ^ 2 + x} + x}{\sqrt{x ^ 2 + x} + x}[/tex] to obtain:
    [tex]\lim_{x \rightarrow +\infty} \sqrt{x ^ 2 + x} - x = \lim_{x \rightarrow +\infty} \frac{(\sqrt{x ^ 2 + x} - x) (\sqrt{x ^ 2 + x} + x)}{\sqrt{x ^ 2 + x} + x}[/tex]
    [tex]= \lim_{x \rightarrow +\infty} \frac{x ^ 2 + x - x ^ 2}{\sqrt{x ^ 2 + x} + x} = \lim_{x \rightarrow +\infty} \frac{x}{\sqrt{x ^ 2 + x} + x}[/tex]
    Now divide both numerator and denominator by x, note that as x tends to positive infinity, then x > 0, so [tex]x = \sqrt{x ^ 2}[/tex]
    [tex]\lim_{x \rightarrow +\infty} \frac{x}{\sqrt{x ^ 2 + x} + x} = \lim_{x \rightarrow +\infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} = \frac{1}{2}[/tex].
    Can you get this? :)
    Do you mean differentiable? Yes, kind of, if f(x0) is differentiable at x0, then we must have:
    [tex]f'(x_0 ^ +) = f'(x_0 ^ -)[/tex].
    By the way, we don't need a slash before an x in your LaTeX code, we just need to type x_0, backslashes are used to make functions displayed in normal font (i.e not in italics) so that they'll stand out from the rest (which is in italic font).
    Example: \sin x
    [tex]\sin x[/tex]
    Last edited: Mar 11, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook