1. ## lim_(x->0^+) (tan(3 x))^x

how would i go about doing this problem?

lim_(x->0^+) (tan(3 x))^x

I am thinking of using lim_(x->0^+) y =?

lny = xln(tan(3x))

lntan(3x)/ (x^-1)

3(sec(3x)^2 / (-x^-2)

am i on the right path?

2. ## Re: lim_(x->0^+) (tan(3 x))^x

derivative of $\ln[\tan(3x)]$ is wrong

3. ## Re: lim_(x->0^+) (tan(3 x))^x

What you're doing is good, write the limit as:
$\lim_{x\to 0^{+}} [\tan(x)]^{x}=\lim_{x\to 0^{+}} e^{x\cdot \ln\left[\tan(x)]}$

Now you can just determine the limit, you don't have to use l'Hopitals rule.

4. ## Re: lim_(x->0^+) (tan(3 x))^x

Out of curiosity: using $\tan \epsilon \sim \epsilon\;(\epsilon\to 0^+)$ we have $\lim_{x\to 0^+}\dfrac{\tan 3x}{3x}=1$ so

$L=\displaystyle\lim_{x \to 0^+}\left(\tan 3x}\right)^x= \displaystyle\lim_{x \to 0^+}\left(\dfrac{\tan 3x}{3x}\cdot 3x\right)^x=$

$\displaystyle\lim_{x \to 0^+}\left(\dfrac{\tan 3x}{3x}\right)^x\cdot \displaystyle\lim_{x \to 0^+}\left( 3x\right)^x= \displaystyle\lim_{x \to 0^+}\left( 3x\right)^x =\ldots$

5. ## Re: lim_(x->0^+) (tan(3 x))^x

Originally Posted by NeoSonata
how would i go about doing this problem?

lim_(x->0^+) (tan(3 x))^x

I am thinking of using lim_(x->0^+) y =?

lny = xln(tan(3x))

lntan(3x)/ (x^-1)

3(sec(3x)^2 / (-x^-2)

am i on the right path?
Yes, You are!... if You consider a more general complex variable z You have...

$\tan 3 z = 3\ z\ \prod_{n=1}^{\infty} \frac{1-\frac{9\ z^{2}}{n^{2} \pi^{2}}}{1-\frac{9\ z^{2}}{(2 n-1)^{2} \pi^{2}}}$ (1)

... and from (1) You derive...

$\ln \tan 3z = \ln 3 + \ln z + \sum_{n=1}^{\infty} \ln \{1-\frac{9\ z^{2}}{n^{2} \pi^{2}}\} - \sum_{n=1}^{\infty} \ln \{1-\frac{9\ z^{2}}{(2 n-1)^{2} \pi^{2}}\}$ (2)

Now from (2) is easy to derive that is...

$\lim_{z \rightarrow 0} z\ \ln \tan 3 z = 0$ (3)

... so that...

$\lim_{z \rightarrow 0} (\tan 3 z)^{z} = 1$ (4)

It is remarkable the fact that the limit (3) and (4) are independent from how z tends to 0...

Kind regards

$\chi$ $\sigma$