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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- Dec 9th 2010, 03:04 PMRColaHow 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? - Dec 9th 2010, 03:25 PMTheEmptySet
Suppose the limit exists and is equal to $\displaystyle L$ then take the natural log of both sides to get

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

Now we can apply L'hosiptials rule

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

Now the limit is $\displaystyle \ln(L)=3 \iff L=e^{3}$ - Dec 9th 2010, 03:48 PMRCola
- Dec 9th 2010, 03:53 PMTheEmptySet
By the chain rule

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

Now by the chain rule again

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

This is 3 function compositions.

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

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

$\displaystyle \displaystyle 2\tan(\sqrt{x})\sec^{2}({\sqrt{x}})\frac{1}{2\

\sqrt{x}}$ - Dec 9th 2010, 03:57 PMProve It
Is it easier if you write the function as

$\displaystyle \displaystyle \ln{L} = \frac{3\ln{(\sec^2{\sqrt{x}})}}{x}$? - Dec 9th 2010, 04:09 PMRCola
- Dec 9th 2010, 04:10 PMProve It
It most certainly is.

$\displaystyle \displaystyle 1 + \tan^2{\theta} = \sec^2{\theta}$. In this case $\displaystyle \displaystyle \theta = \sqrt{x}$. - Dec 9th 2010, 04:13 PMRCola
**Thanks a lot for your help guys!**