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Math Help - Limit Problem

  1. #1
    Member McScruffy's Avatar
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    Limit Problem

    It looks like it should be simple, but I'm having some trouble wrapping my head around this one. Any help would be greatly appreciated.

    \lim_{x\rightarrow3}\bigg(\frac{x}{x-3}\int_{3}^{x}\frac{\sin{t}}{t} dt\bigg)

    Thanks.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by McScruffy View Post
    It looks like it should be simple, but I'm having some trouble wrapping my head around this one. Any help would be greatly appreciated.

    \lim_{x\rightarrow3}\bigg(\frac{x}{x-3}\int_{3}^{x}\frac{\sin{t}}{t} dt\bigg)

    Thanks.
    Using the usual algebra of limits you have \left( \lim_{x\rightarrow3} x\right) \cdot \lim_{x \rightarrow 3} \frac{\int_{3}^{x}\frac{\sin{t}}{t} dt}{x - 3}.

    The second limit has the indeterminant form 0/0 so apply l'Hopitals to it (use the Fundamental Theorem of Calculus to differentiate the numerator).
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  3. #3
    Member McScruffy's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Using the usual algebra of limits you have \left( \lim_{x\rightarrow3} x\right) \cdot \lim_{x \rightarrow 3} \frac{\int_{3}^{x}\frac{\sin{t}}{t} dt}{x - 3}.

    The second limit has the indeterminant form 0/0 so apply l'Hopitals to it (use the Fundamental Theorem of Calculus to differentiate the numerator).
    Got it, thanks.
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by McScruffy View Post
    It looks like it should be simple, but I'm having some trouble wrapping my head around this one. Any help would be greatly appreciated.

    \lim_{x\rightarrow3}\bigg(\frac{x}{x-3}\int_{3}^{x}\frac{\sin{t}}{t} dt\bigg)

    Thanks.
    Perhaps you could write \int_{3}^{x}\frac{\sin{t}}{t} dt as a function of x.

    That is F(x)=\int_{3}^{x}\frac{\sin{t}}{t}dt

    Not really sure how to proceed though...
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  5. #5
    Super Member Random Variable's Avatar
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    Let F(t) be an antiderivative of  \frac{\sin t}{t}

    then  \lim_{x \to 3} \frac{x}{x-3} \int^{x}_{3} \frac{\sin t}{t} \ dt = \lim_{x \to 3} \frac{x}{x-3} \big(F(x)-F(3)\big)

     = \lim_{x \to 3} x \cdot \lim_{x \to 3} \frac{F(x)-F(3)}{x-3}

     = 3F'(3) = 3 \ \frac{\sin 3}{3} = \sin 3
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Random Variable View Post
    Let F(t) be an antiderivative of  \frac{\sin t}{t}

    then  \lim_{x \to 3} \frac{x}{x-3} \int^{x}_{3} \frac{\sin t}{t} \ dt = \lim_{x \to 3} \frac{x}{x-3} \big(F(x)-F(3)\big)

     = \lim_{x \to 3} x \cdot \lim_{x \to 3} \frac{F(x)-F(3)}{x-3}

     = 3F'(3) = 3 \ \frac{\sin 3}{3} = \sin 3
    That was really nice, man.
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