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 colin@flyingcoloursmaths.co.uk 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

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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