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Homework Help: Limits! (one algebreic, two Trig)

  1. Sep 13, 2004 #1
    Well I just started to learn limits in my calculus class, and I'm getting it fairly well, but I'm a bit confused on these three:

    1) the limit of (x^5 - 32)/(x-2) as x approaches 2.
    The problem I have here is factoring the (x^5 - 32) part. Once I can get that factored I think I can handle the rest. :D

    2) what is the limit of xsecx as x appraoches pi.
    I simplified that to x(1/cosx) but that really got me nowhere.

    3) and, last but not least, what is the limit of (1-tanx)/(sinx-cosx) as x approaches pi/4?
    I broke this one down to [(2 sin 2x)/2x)(mx/3 sin 3x)], but I don't know where to go from there.

    The two basic trig fxns I learned are:
    the limit of sin x/x as x approaches 0 is equal to one
    the limit of (1-cos x)/x as x approaches 0 is equal to zero

    If anybody coiuld please lend me their wisdom, lead me into the right direction, I would be forever grateful. Thank you!
    Last edited: Sep 13, 2004
  2. jcsd
  3. Sep 13, 2004 #2


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    Did you try substituting for 2 before trying to transform the expression, you must first try and see if it will create an inderterminate form, i see 0/16 which is 0
  4. Sep 13, 2004 #3
    YEah, I typed the problem wrong, but I edited it. And yes I did plug in, and got 0/0 so now that means I need to factor further, so... heh.
  5. Sep 13, 2004 #4


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    You might recognize (y^n - q)/(y-1) as the sum of a geometric series and go from there. If not, try dividing x - 2 directly into x^5 - 32. You should see a pattern emerge even before you finish it! :-)
  6. Sep 14, 2004 #5
    Have you not learned the Product Rule and Quotient Rule? If you know these two rules, solving the problems you posted will only be a matter of plugging in the value of x.
  7. Sep 14, 2004 #6


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    No, it's not. Problem 2 can be handled that way, but problems 1 and 3 both have denominators that are 0 at the "target" point and so the quotient rule does not apply (in fact, it doesn't apply in all the interesting cases!).

    Alethia was correct- you need to factor (x5- 32).

    In general, xn-an can be factored as
    (x-a)(xn-1+ axn-2+ a2xn-3+...+an-2x+ an-1).
    x5- 32= (x-2)(x4+ 2x3+ 4x2+ 8x+ 16).

    I'm not at all sure how Alethia got the "3x" and "2x" in the 3rd problem (nor what "m" is!). Since tan(x)= sin(x)/cos(x), 1- tan(x)= 1- sin(x)/cos(x)=
    (cos(x)- sin(x))/cos(x) so (for x not &pi/4)

    (1- tan(x))/(sin(x)- cos(x))= ((cos(x)- sin(x))/cos(x))(1/(sin(x)- cos(x)) and that should be easy.
  8. Sep 14, 2004 #7
    Thank you everyone! I think I just factored wrong on the first one. I used synthetic, but I kept getting the wrong answer, but I get it now. :D
  9. Sep 14, 2004 #8
    Woops, sorry for the blunder I made. I'll try to be more careful in the future.
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