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Homework Help: Finding Limits (w/o calc)

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    I need to figure out these limits without using a graphing calculator:




    3. The attempt at a solution

    I really had no idea about this one.
    Maybe there's an identity I'm forgetting about.

    For this one, I multiplied the top and bottom by the conjugate of the top, which was
    sqrt(x+4) + 2. But it backfired because I ended up with 0/0 as my limit again which
    doesn't exist.

    For this one, I used a table of values. I got the right answer, but I'm wondering if there's
    any way to do it algebraically.

    I would be so thankful for any help at all on these ^_^ o:)

  2. jcsd
  3. Oct 13, 2009 #2


    Staff: Mentor

    I don't have any ideas for #1, but here are some tips for 2 and 3.
    2. You had the right idea, multiplying by the conjugate. After you multiply, you should end up with only x in the numerator, which cancels with the x in the denominator.
    3. I haven't worked this through, but both numerator and denominator have factors of (x - 1). Rewrite both numerator and denominator as (x - 1)* (the rest), cancel the x - 1 factors, and you should be able to evaluate the limit.
  4. Oct 13, 2009 #3
    For the first one notice that [tex] \frac{sin(x)}{x + tan(x)} = \frac{sin(x)}{x(1 + \frac{tan(x)}{x})} [/tex].

    [tex] \frac{tan(x)}{x} = \frac{\frac{sin(x)}{cos(x)}}{x} = \frac{sin(x)}{xcos(x)}[/tex]

    So [tex] \frac{sin(x)}{x + tan(x)} = \frac{sin(x)}{x} \left \left \left \frac{1}{1 + \frac{sin(x)}{x} + \frac{1}{cos(x)} } [/tex].

    You know the limit of sin(x)/x as x goes to 0, right? So pass to the limit now, using limit laws.
  5. Oct 13, 2009 #4
    I messed up the last equation in my previous post. It should be:

    [tex] \frac{sin(x)}{x + tan(x)} = \frac{sin(x)}{x} \left \left \left \frac{1}{1 + \frac{sin(x)}{x} \left \left \frac{1}{cos(x)} } [/tex]
  6. Oct 13, 2009 #5
    Thanks guys

    JG89 actually I don't know the limit of Sinx/x as x goes to 0 anymore. I'm sure I did at one time. ... what is it again?
  7. Oct 13, 2009 #6
    It's 1
  8. Oct 13, 2009 #7
    Oookay. that's smart
    it ends up being 1/2, which is the same as the answer key.
  9. Oct 13, 2009 #8
    Also just to add a bit more, I believe you can use L'Hopital's Rule when you have 0/0 or infinity/infinity.
  10. Oct 14, 2009 #9


    User Avatar
    Science Advisor

    Yes, but the title of this thread is "Finding Limits (w/0 calc)".
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