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Math Help - Limit problem

  1. #1
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    Limit problem

    I need to found a limit, x->0 of (e^x + sinx)^ctgx, but without L'Hospital rule. I know how to get it with L'H, and I see it is 1^inf form, and i need to get it on form 1 + somthing, but how ? Any ideas?
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  2. #2
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    Re: Limit problem

    Quote Originally Posted by DonnieDarko View Post
    I need to found a limit, x->0 of (e^x + sinx)^ctgx, but without L'Hospital rule. I know how to get it with L'H, and I see it is 1^inf form, and i need to get it on form 1 + somthing, but how ? Any ideas?
    What is ctgx?
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  3. #3
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    Re: Limit problem

    cot(x), or cotangent function. 1/tan(x)
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    Re: Limit problem

    \displaystyle \begin{align*} \lim_{x \to 0}\left(e^x + \sin{x}\right)^{\cot{x}} &= \lim_{x \to 0}e^{\ln{\left[ \left( e^x + \sin{x} \right)^{\cot{x}} \right]}} \\ &= \lim_{x \to 0}e^{\cot{x}\ln{\left( e^x + \sin{x} \right)}} \\ &= \lim_{x \to 0}e^{\frac{\cos{x}\ln{\left( e^x + \sin{x} \right)}}{\sin{x}}} \\ &= e^{\lim_{x \to 0}\frac{\cos{x}\ln{\left( e^x + \sin{x} \right)}}{\sin{x}}} \end{align*}

    This is now in \displaystyle \begin{align*} \frac{0}{0} \end{align*} form, so you can apply L'Hospital's Rule.
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  5. #5
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    Re: Limit problem

    I know that, but I must solve it without applying L'Hospital's Rule anywhere.
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  6. #6
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    Re: Limit problem

    Solved. Here's the solution.

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