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 $\displaystyle L$ 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 $\displaystyle \ln(L)=3 \iff L=e^{3}$
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
$\displaystyle \displaystyle 2\tan(\sqrt{x})\sec^{2}({\sqrt{x}})\frac{1}{2\
\sqrt{x}}$