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BC Calculus help

  1. Jul 1, 2005 #1
    Sorry I'm not very good at using latex, but here's my shot.

    1. Let f(x) = sqrt (1+2x) - 1 - sqrt (x). Find some a where a is positive, such that lim of [tex]\frac{f(x)}{x^a}[/tex] as x approaches 0 from the right is finite and non zero.

    I know the problem requires the use of L'Hopital's rule, but I seem to be making circles. When I differentiate, I keep getting indeterminite forms and can't get the limit to be finite

    2. A picture 1.4 meters high stands on a wall so that its lower edge is 1.8 meters above the eye of an observer. What is the most favorable distance from the wall for this observer too stand - that is, to maximize his or her angle of vision.

    I started by doing tan(theta) = [tex]\frac{1.4+1.8}{x}[/tex] where x is the distance from the wall. To maximize the angle, I took the arctan of each side and differentiated, looking for when it will equal 0, but the derivative is never 0.

    3. Evaluate for any fixed number k>0:

    lim [tex]\frac{(1^k + 2^k + ... n^k)}{n^(k+1)}[/tex] as n approaches infinity
    [Edit: The denominator should be n raised to the quantity of (k+1)]

    The numerator looks like a Riemann sum, but I have no idea how to begin solving it.

    Thanks for any help
    Last edited: Jul 1, 2005
  2. jcsd
  3. Jul 1, 2005 #2


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    For number 1, there is an obvious choice of a to try that might work even without L'Hospital's rule.

    For number 2, I think you need to try law of cosines applied to the triangle from the eye to the bottom and top edges of the picture.

    For number 3, look at it term by term. The largest term in the numerator is the last one. What do you get when you divide it by the denominator? What about all the terms in front of that one.

    Hint: Write the denominator as

    [tex] nn^k [/tex]
    Last edited: Jul 1, 2005
  4. Jul 4, 2005 #3
    Thanks OlderDan, I solved the first two problems, but I'm still having trouble with the third. If I seperate the numbers into a series of fractions, like

    1^k over n^k+1 plus 2^k over n^(k+1) plus 3^k over n^(k+1) and end with n^k over n^(k+1), would that give me the answer (which would be 0?)

  5. Jul 4, 2005 #4


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    Looks like you about have it. Every term can be written as (1/n) times a factor of 1 or less, so each goes to zero in the limit of infinite n. With an infinite nubmer of terms in the series, you should do a bit more to prove that the sum of all these terms is zero. You could do that by factoring out 1/n from every term and proving that the infinite series that remains still converges by one of the standard tests for convergence, or by applying such tests to the series as it stands.
  6. Jul 5, 2005 #5
    Thanks OlderDan, I think I have all three of the problems solved now.
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