*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I have a problem with a limit! I need to figure out what $\left( \tan \left(x\right) \right)^x$ is as $x \rightarrow 0$.

-- Brilliant Explanation Required Now! Our Understanding's Limited; L'Hôpital's Inept

Right, BERNOULLI, stop badmouthing L'Hôpital and let's figure out this limit. It's clearly an indeterminate form to begin with, as as $x$ gets small, $(\tan(x))^x$ approaches $0^0$, which isn't defined.

So let's do something clever. If we call the limit $L$ and take logs, we get:

$\ln(L) = x \ln ( \tan(x) )$, which is again indeterminate -- although now it's (loosely) $0 \times \infty$, which isn't defined.

I'm going to use one of my favourite tricks, which is to rewrite $\tan(x)$ as $ x \times \frac{\tan(x)}{x}$, making the log of the limit:

$\ln(L) = x \ln (x) + x \ln\left( \frac{\tan(x)}{x} \right)$

Applying L'Hôpital's rule to $\frac{\tan(x)}{x}$ gives a limit as $x \rightarrow 0$ of $\frac{\sec(x)}{1}$, which goes to 1, meaning that the second term goes to $0 \times 0$ and can be ignored.

The first term can be rewritten as $\frac{\ln(x)}{1/x}$, another indeterminate form we can apply L'Hôpital to -- the limit becomes $\frac{\frac{1}{x}}{\frac{-1}{x^2}} = -x$, which again goes to 0 as $x \rightarrow 0$.

So, in the limit as $x \rightarrow 0$, $\ln(L) \rightarrow 0$, and as a result, $L \rightarrow 1$.

Hope that helps!

-- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.