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Math Help - How to solve limit of the function?

  1. #1
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    How to solve limit of the function?

    How to solve limit of the function?

    Limit[(1 + (Tan[Sqrt[x]])^2)^(3/x), x -> 0]
    I'm getting answer from Math program = e^3

    How do we get there?
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  2. #2
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    Quote Originally Posted by RCola View Post
    How to solve limit of the function?

    Limit[(1 + (Tan[Sqrt[x]])^2)^(3/x), x -> 0]
    I'm getting answer from Math program = e^3

    How do we get there?
    Suppose the limit exists and is equal to L then take the natural log of both sides to get

    \displaystyle \ln(L)=\frac{3}{x}\ln\left(1+\tan^2(\sqrt{x}) \right)

    Now we can apply L'hosiptials rule

    \displaystyle 3\frac{\frac{2\tan(\sqrt{x})\sec^{2}(\sqrt{x})\cdo  t \frac{1}{2\sqrt{x}}}{1+\tan^2(\sqrt{x})}}{1}=3\fra  c{\tan(\sqrt{x})}{\sqrt{x}}=3

    Now the limit is \ln(L)=3 \iff L=e^{3}
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post

    \displaystyle {\frac{2\tan(\sqrt{x})\sec^{2}(\sqrt{x})\cdot \frac{1}{2\sqrt{x}}}}
    I don't understand why this expression is here.

    (ln|x|)' = 1/x

    Maybe it's composite function. So how do we get this line?
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  4. #4
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    Quote Originally Posted by RCola View Post
    I don't understand why this expression is here.

    (ln|x|)' = 1/x

    Maybe it's composite function. So how do we get this line?
    By the chain rule

    \frac{d}{dx} \ln(g(x))=\frac{1}{g(x)}\cdot g'(x)

    Now by the chain rule again

    \displaystle \frac{d}{dx} 1+\tan^2(\sqrt{x})

    This is 3 function compositions.

    f(t)=t^2;t=\tan(w);w=\sqrt{x}

    \displaystyle \frac{df}{dt}\frac{dt}{dw}\frac{dw}{dx}=2t\sec^2{w  }\frac{1}{2x} Now just sub in to get

    \displaystyle 2\tan(\sqrt{x})\sec^{2}({\sqrt{x}})\frac{1}{2\<br />
\sqrt{x}}
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  5. #5
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    Is it easier if you write the function as

    \displaystyle \ln{L} = \frac{3\ln{(\sec^2{\sqrt{x}})}}{x}?
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  6. #6
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    Quote Originally Posted by Prove It View Post
    Is it easier if you write the function as

    \displaystyle \ln{L} = \frac{3\ln{(\sec^2{\sqrt{x}})}}{x}?
    Is \displaystyle(1+\tan^2(\sqrt{x}) equals to (\sec^2{\sqrt{x}}) ?
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  7. #7
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    It most certainly is.

    \displaystyle 1 + \tan^2{\theta} = \sec^2{\theta}. In this case \displaystyle \theta = \sqrt{x}.
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  8. #8
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    Thanks a lot for your help guys!
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