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Posted by on Sunday, February 17, 2008 at 7:28pm.

Below are the 5 problems which I had trouble in. I can't seem to get the answer in the back of the book. Thanks for the help!

lim (theta-pi/2)sec(theta)
theta->pi/2
Answer: -1
I am not sure what to do here.

lim (tan(theta))^(theta)
theta->0+
Answer:1
ln(tan(theta))/(1/theta)
L'Hospital's Rule:
lim [(1/(tan(theta))(sec^2(theta))]/(-theta->0+ /x^2]

lim (theta)^2(sin(1/(theta))
theta->oo
Answer:oo
lim (1/h)^2(sin(h))
h->0+
L'Hospital's Rule:
cos(h)/(-2/h)

lim (csc^-1 x)/1/x
x->oo
Answer: 1
I am not sure how 1 can be the answer.

integral 0 to 3 dx/sqrt(9-x^2)
Answer: pi/2
I know this is some form of arcsin(x)->f'(x)=1/sqrt(1-x^2), but I don't know what to do with the 9.

  • Calculus - , Sunday, February 17, 2008 at 7:44pm

    lim (theta-pi/2)sec(theta)
    theta->pi/2
    Answer: -1
    I am not sure what to do here.
    ++++++++++++++++
    (T-pi/2)/cos(T)
    cos T = sin(pi/2 - T)
    so we have
    -(pi/2 -T) / sin (pi/2-T)
    as T ---> pi/2 we have sin of small angle and sin(angle) ---->angle + small stuff series in angle^3/3! etc
    so in the limit we have
    -(pi/2-T)/(pi/2 - T) = 1

  • Without series - , Sunday, February 17, 2008 at 8:42pm

    lim (theta-pi/2)sec(theta)
    theta->pi/2
    Answer: -1
    I am not sure what to do here.
    ++++++++++++++++
    numerator = (T-pi/2)
    denominator = cos (T)
    d/dT numerator = 1
    d/dT denominator = - sin T
    as T -->pi/2, sin T ---> 1
    so
    1/-1
    = -1

  • Without series - , Sunday, February 17, 2008 at 8:46pm

    thanks

  • Calculus - , Sunday, February 17, 2008 at 7:45pm

    Sorry
    -1 above

  • Calculus - , Sunday, February 17, 2008 at 7:56pm

    lim (tan(theta))^(theta)
    theta->0+
    Answer:1
    ln(tan(theta))/(1/theta)
    L'Hospital's Rule:
    lim [(1/(tan(theta))(sec^2(theta))]/(-theta->0+ /x^2]
    ============================
    (tan T)^T as T--->0+
    log answer = T ln tan T
    log answer = T ln ( T + small stuff)
    log answer = T (T-1) + small stuff
    log answer = -1 T --->0
    e^0 = 1

  • without series - what a mess - , Sunday, February 17, 2008 at 9:02pm

    My way sure is easier !

    ln answer = T ln tan T
    = ln tan T/1/T

    numerator = ln tan T
    denominator = 1/T

    d/dT numerator = (cos T/sin T)(1/cos^2 T)
    = 1/(cos T sin T)
    but
    sin T cos T = .5 sin 2T

    d/dT denominator = -1/T^2

    as T --> 0 sin 2T ---> 2 T
    so
    we have
    1/T / -1/T^2 = -T ---> 0
    e^0 = 1

  • Calculus - , Sunday, February 17, 2008 at 7:58pm

    I am seeing a pattern here. Use a Taylor series expansion for your functions.

  • Calculus - , Sunday, February 17, 2008 at 8:03pm

    lim (theta)^2(sin(1/(theta))
    theta->oo
    Answer:oo
    lim (1/h)^2(sin(h))
    h->0+
    L'Hospital's Rule:
    cos(h)/(-2/h)
    ++++++++++++++++++++++++++++++++++=
    T^2 sin(1/T) as T--> oo
    again sin A ---> A -A^3/3!..... for small angle
    so we have
    T^2/ (1/T) = T ^3 for big T
    that gets very big as T gets moderately large.

  • Calculus - , Sunday, February 17, 2008 at 9:12pm

    numerator = sin (1/T)
    denominator = 1/T^2

    d/dT numerator = -(1/T^2 )cos (1/T)
    d/dt denominator = -2T/T^4 = - 2/T^3

    ratio = 2 T cos (1/T)
    cos 0 = 1
    so
    ratio = 2 T which is oo as T ---> oo

  • Calculus - , Sunday, February 17, 2008 at 8:14pm

    lim (csc^-1 x)/1/x
    x->oo
    Answer: 1
    I am not sure how 1 can be the answer.
    ==========================
    numerator = csc^-1 x
    denominator = (1/x)
    d/dx numerator = -1/x sqrt(x^2-1)
    d/dx denominator = -1/x^2

    as x gets big we have -1/x^2 for the numerator and -1/x^2 for the denominator

  • Calculus - , Sunday, February 17, 2008 at 8:19pm

    integral 0 to 3 dx/sqrt(9-x^2)
    Answer: pi/2
    I know this is some form of arcsin(x)->f'(x)=1/sqrt(1-x^2), but I don't know what to do with the 9.
    ================================
    let x = 3 sin t
    dx = 3 cos t dt
    so
    3 cos t dt /sqrt(9 - 9 sin^2 t)
    3 cos t dt / 3 sqrt (1-sin^2 t)
    but 1 - sin^2 = cos ^2
    You can take it from there.

  • Calculus (for damon) - , Sunday, February 17, 2008 at 8:23pm

    about the taylor expansion. I haven't learned that yet. Is there an alternative way?

  • Calculus (for damon) - , Sunday, February 17, 2008 at 8:34pm

    Have you had any series even in algebra 2 ?

  • Calculus - , Sunday, February 17, 2008 at 8:39pm

    yes, but that is a long time ago

  • Calculus - , Sunday, February 17, 2008 at 9:08pm

    N-th order Taylor expansion:

    f(x+h) = f(x) + h f'(x) + h^2/2 f''(x) +h^3/3! f'''(x) + ...+ h^N/N! f^(N)(x) +
    O(h^N+1)

    Assuming that f is continuously differentiable N times. O(h^(N+1)) means a term proportional to h^(N+1) in the limit h ---> 0.

    A non-rigorous proof goes as follows. Just assume that f(x+h) can be expressed as a power series in h:

    f(x + h) = A + B h + C h^2 + ...

    Then take the limit h --->0 of both sides to find A. Take the first derivative and the limit h ---> 0 to find B, the second derivative and the limit h ---> 0 to find C, etc.

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