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Math Help - limits using L'Hopital's rule or otherwise and using leading behaviours to sketch

  1. #1
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    limits using L'Hopital's rule or otherwise and using leading behaviours to sketch

    Hi, I tried to do the following problems but I didn't get very far.

    Find each limit using L'Hopital's rule (if possible; check assumptions before you try to use it!) or otherwise.

    lim (tanx/x) as x approaches 0
    -->since it's indeterminate form : 0/0, i used the rule and got 0 but the answer is 1.

    lim (x^(4x)) as x approaches 0 from the right side
    -->i tried to use the rule for it by changing it into a fraction: 1/ x^(-4x) but then when i differentiate it's almost the same thing

    a(c) = 5c / e^(2c)
    -->how do i find the leading behaviours if there is only one term in the numerator and in the denominator?

    thanks!
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  2. #2
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    Re: limits using L'Hopital's rule or otherwise and using leading behaviours to sketc

    \lim_{x \to 0}\frac{\tan x}{x}=\lim_{x \to 0}\frac{\sec ^{2}x}{1}=1
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  3. #3
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    Re: limits using L'Hopital's rule or otherwise and using leading behaviours to sketc

    \lim_{x \to 0+}x^{4x}=\lim_{x \to 0+}e^{ln(x)4x}
    =e^{\lim_{x \to 0+}ln(x)4x}
    =e^{4\lim_{x \to 0+}\frac{ln(x)}{\frac{1}{x}}}
    =1
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: limits using L'Hopital's rule or otherwise and using leading behaviours to sketc

    An other way to solve the first limit:
    \lim_{x \to 0} \frac{\tan(x)}{x}=\lim_{x \to 0} \frac{\frac{\sin(x)}{\cos(x)}}{x}=\lim_{x\to 0} \left(\frac{\sin(x)}{x}\cdot \frac{1}{\cos(x)}\right)=\lim_{x\to 0} \frac{\sin(x)}{x} \cdot \lim_{x \to 0}\frac{1}{\cos(x)}=1.1=1
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