Ask Uncle Colin: another vile limit

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 and Uncle Colin will do what he can.

Dear Uncle Colin,

Apparently, you can use L'Hôpital's rule to find the limit of $\left(\tan(x)\right)^x$ as $x$ goes to 0 - but I can't see how!

- Fractions Required, Example Given Excepted

Hi, FREGE, and thanks for your question!

As it stands, you can't use L'Hôpital - but you can adjust it so you can!

If you take logs, you get $x \ln( \tan(x))$, which you can rewrite as $\frac{\ln(\tan(x))}{\frac{1}{x}}$.

Now, that's not the nicest of things to compute, but it's doable. A quick check that it's undefined on the top and the bottom (it is!) and we're away.

The top differentiates to $\frac{\sec^2(x)}{\tan(x)}$ and the bottom to $-\frac{1}{x^2}$, so our limit is now $-\frac{x^2 \sec^2(x)}{ \tan(x)}$ - but I'm sure we can make that a bit less horrible.

Let's turn it into $-\frac{x^2}{ \sin(x)\cos(x)}$, which is already a lot nicer. It's still undefined, but it's nicer; it'll be nicer still when we replace $\sin(x)\cos(x)$ with $\frac{1}{2}\sin(2x)$.

So we now have $\frac{-2x^2}{\sin(2x)}$ as our limit. Differentiating the top gives $-4x$, and the bottom goes to $2\cos(2x)$ and finally we have something we can calculate: we get 0.

However, that's the logarithm of the limit we wanted: our final answer is $e^0 = 1$.

Hope that helps!

-- Uncle 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.


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