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Indertiminate form

  1. Mar 24, 2004 #1
    we starting to learn about l'hospital's rule and i can't figure out this problem. It goes:

    Find the Limit for the following indeterminate forms of the type "infinity - infinity"

    limit as x approaches 0 of [(1/sin x)-(1/x)]
  2. jcsd
  3. Mar 24, 2004 #2
    God it's been too long and my memory of particular derivatives is fading, but if you find a common denominator of [tex]x sin(x)[/tex] because:

    [tex]\frac{1}{sin(x)}-\frac{1}{x} = \frac{x-sin(x)}{x \ sin(x)}[/tex]

    and we call the fraction


    can you find the derivatives [tex]f'(x)[/tex] and [tex]g'(x)[/tex] to find the limit of the original form? You'll have to use the chain rule, right? :wink: Like I said it has been too long...hehe

    oops, that product rule, not chain rule!
    Last edited: Mar 24, 2004
  4. Mar 24, 2004 #3
    Okay ACLerok I couldn't help myself and gave it my best shot. See if you get the same thing:


    [tex]\frac{x-sin(x)}{x \ sin(x)} = \frac{f(x)}{g(x)}[/tex]


    [tex]f'(x) = 1-cos(x)[/tex]


    [tex]g'(x) = sin(x) + x \ cos(x)[/tex]

    so the format is now

    [tex]\lim_{x \rightarrow 0} \ \frac{1-cos(x)}{sin(x)+x \ cos(x)} = \frac{1-1}{0+(0)(1)}[/tex]

    that's indeterminate, too, so take the drivative again to get

    [tex]\lim_{x \rightarrow 0} \ \frac{sin(x)}{2cos(x)-x \ sin(x)}[/tex]

    after some algebraic hoopla you get

    [tex]\lim_{x \rightarrow 0} \ \frac{1}{\frac{2cos(x)}{sin(x)}-x} = \frac{1}{\infty - 0} = 0[/tex]

    So the limit is zero.

    NOTE: I updated this page once to correct errors I made during the computation, starting with the SECOND derivative
    Last edited: Mar 24, 2004
  5. Mar 25, 2004 #4
    Hi and thanks alot for replying. I understand everything you did but one thing. can you please explain why the limit of 2cos(x)/sin(x) as x approaches 0 equals infinity?
  6. Mar 25, 2004 #5
    2cos(x)/sin(x) is 2/tan(x). tan(0) is 0, so 2cos(x)/sin(x) is 2/0.

  7. Mar 25, 2004 #6
    Cookiemonster said it all, but I wanted to add that an easy way to spot this sort of thing is punch sin(0.0000001) in the calc. You get a VERY small number, and if you punched a smaller number than 0.0000001 you'd get an even smaller number for the sin() value. Ad infinitum. If you divide by a fraction you multiply the numerator.

    As long as the numerator is a positive value (no matter how big or how small) you get positive infinity for 1/sin(x) as x -> 0
  8. Mar 25, 2004 #7


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    Homework Helper

    Make sure your calculator is in radians and you should see a moderately interesting result (which, incidently, supports the limit going to zero in the original difference expression).
  9. Mar 25, 2004 #8
    isn't 2/0 undefined?
  10. Mar 25, 2004 #9
    Yes, but lim{x->0+} 2/x is infinity and lim{x->0-} 2/x is -infinity, so either way it goes to a big number.

    Edit: I should point out that tan(x) is continuous in the neighborhood of 0, so that's why it goes to a big number.

    Last edited: Mar 26, 2004
  11. Mar 26, 2004 #10
    Right. ACLerok keep in mind that when referring to limits as x -> 0 people tend to refer to fractions like



    2/(an arbitrarily tiny number that's so close to zero the numerator is infinity)

    Thanks for this thread, btw! It was fun to work through.
  12. Mar 27, 2004 #11
    ok thanks alot guys! one down two more tough ones to go...
  13. Mar 28, 2004 #12
    click here for problem

    is this problem exactly the same as the first one i posted or is there something i have to do first in order to differentiate?
  14. Mar 28, 2004 #13
    Try putting the two terms over one denominator and simplifying.

  15. Mar 29, 2004 #14
    Did you figure that last one out AC?
  16. Mar 29, 2004 #15
    no i actually just looked at a friend's answer.

    this is the second problem i am given.

    click here for 2nd problem

    after simplifying i get lim x->0, (1-x/x^2)

    i figure that the lim x->0, (1/x^2), and x^2 being a very small number, is +infinity.

    am i right or wrong?
    Last edited: Mar 29, 2004
  17. Mar 29, 2004 #16
    I think you should not have given up so easily. The problem looks much worse than it actually is.

    [tex] lim \ x \rightarrow 0 \ (\frac{1+x}{x}-\frac{1-x}{x})[/tex]

    [tex] = \frac{1+x-(1-x)}{x} = \frac{1+x-1+x}{x} = \frac{2x}{x} = 2[/tex]
    Last edited: Mar 29, 2004
  18. Mar 29, 2004 #17
    I like it
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