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Math Help - test questions

  1. #1
    Senior Member polymerase's Avatar
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    test questions

    Hey,

    I have a calculus test coming up (first year) and i was wondering if you guys would happen to know of any questions (HARD ONES but still within the realm of first year) that i can practice with. Topics covered will be: limits, derivative, continuity (ie. finding constants a,b,c...such that the 2,3,4.... functions are continuous everywhere), trig and trig derivative, orthogonal trajectories, implicit. My prof always has a "very difficult" question (usually limits or continuity) where you either get 10/10 for correct answer or 0/10 nothing in between. So i need some hard questions to practice with....

    Thanks!
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  2. #2
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    Krizalid's Avatar
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    Okay.

    Here's a "hard limit"

    \lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{\sin\dfrac1x}

    It's a nice one, and of course, it's harder without L'H˘pital's Rule.

    I have a very hard one, but I'll post it when you've solved this one.
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    Okay.

    Here's a "hard limit"

    \lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{\sin\dfrac1x}

    It's a nice one, and of course, it's harder without L'H˘pital's Rule.

    I have a very hard one, but I'll post it when you've solved this one.
    alright ill work on it n let u know. thanks...anything on continuity or the rest? i have an example posted before, its titled "very difficult" piece function question and it was last answerd by galactus. its on pg 3 now.
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    \lim_{x\to\infty}\frac{\dfrac\pi2-\arctan x}{\sin\dfrac1x}
    (This is too hard for a Calculus test).

    \lim_{x\to \infty} \frac{\frac{\pi}{2} - \tan^{-1} x}{\sin \frac{1}{x}} = \lim_{x\to 0} \frac{\frac{\pi}{2} - \tan^{-1} \frac{1}{x}}{\sin x}

    Use the identity that \tan^{-1} \frac{1}{x} = \cot^{-1} x.

    So we get,
    \lim_{x\to 0}\frac{\frac{\pi}{2} - \cot^{-1} x}{\sin x}

    Now use the identity that \boxed{ \frac{\pi}{2} - \cot^{-1} x = \tan^{-1} x } \mbox{ for } x>0

    We get,
    \lim_{x\to 0^+} \frac{\tan^{-1} x}{\sin x} = \lim_{x\to 0^+} \frac{\tan^{-1} x}{x} \cdot \frac{x}{\sin x} = 1\cdot 1 = 1

    Now do the part with x\to 0^-.

    EDIT: Sophomore's Dream with x^{-x} is cooler.

    EDIT2: No need to do x\to 0^-! (Because we are doing the limit x\to \infty.)
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  5. #5
    Math Engineering Student
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    Well... I said it's not a hard limit, but the idea was that polymerase answer this question
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