Thread: How to solve limit of the function?

1. 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?

2. Originally Posted by RCola
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 $\ln(L)=3 \iff L=e^{3}$

3. Originally Posted by TheEmptySet

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

4. Originally Posted by RCola
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

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

5. Is it easier if you write the function as

$\displaystyle \ln{L} = \frac{3\ln{(\sec^2{\sqrt{x}})}}{x}$?

6. Originally Posted by Prove It
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 $(\sec^2{\sqrt{x}})$ ?

7. It most certainly is.

$\displaystyle 1 + \tan^2{\theta} = \sec^2{\theta}$. In this case $\displaystyle \theta = \sqrt{x}$.

8. Thanks a lot for your help guys!