Written by Colin+ in ask uncle colin.

*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

Help! An emergency has arisen and my L'Hôpital's Rule is out of order -- but I still need to find $\lim_{x \to 0_+} \frac{\ln(x)}{\cot(x)}$! This is not a drill!

-- Help! Inexplicable Limitations Bring Everyone Ridiculous Trouble

Don't panic, HILBERT! Don't panic.

Now, for the benefit of anyone who's not constrained by HILBERT's inability to use L'Hôpital's rule1, that is the obvious way to solve it. When

$L = \lim_{x \to x_0} \frac{u}{v}$ is undefined (because $u = 0$ and $v=0$ or $u = \infty$ and $v=\infty$), then $L = \lim_{x \to x_0} \frac{\diff ux}{ \diff vx}$.

In this case,

$\lim_{x \to 0_+} \frac{\ln(x)}{\cot(x)} = \lim_{x \to 0_+} \frac{\frac 1x }{-\cosec^2(x)}= \lim_{x \to 0_+} \frac{ -\sin^2 (x)}{x} $.

Since for all $x > 0$, $\sin(x) < x$, $0 < \sin^2(x) < x^2$. That means $0 < \frac{\sin^2(x)}{x} < x$, so as $x \to 0_+$, $L \to 0$. Sandwich rule for the win!

However, as HILBERT says, L'Hôpital's rule isn't allowed in this urgent problem. So how do we approach it?

The first thing is to rewrite the limit as:

$\lim_{x \to 0_+} {\ln(x)}{\tan(x)}$. It already looks nicer.

We can apply a similar sort of reasoning as before using the sandwich rule, but it needs a bit of setting up: it turns out that $ -\frac 1 {\sqrt{x}} < \ln(x) ≤ 0$ for $ 0 < x ≤ 1$. The upper end of the inequality is obvious; the lower end not so much, so let's show it: it's equivalent to showing $-1 < \sqrt{x} \ln(x)$ in this range. Differentiating $ y= x^{\frac 12} \ln(x)$ gives $\dydx = \frac 12 x^{-\frac 12} \ln(x) + x^{\frac 12} \cdot \frac 1x$. This vanishes when $\frac 1{\sqrt x} \left( \ln(x) + 2 \right) = 0$, or $x = e^{-2}$, at which point $y > -1$. (It works out to be -0.736 or so). That's the only turning point, and it's a global minimum on the domain where it's defined, so the inequality holds on $0 < x ≤ 1$. Lucky! So, we know that $ -\frac 1 {\sqrt{x}} < \ln(x) ≤ 0$ for $ 0 < x ≤ 1$, which means $-\frac{\tan(x)}{\sqrt{x}} < \ln(x)\tan(x) ≤ 0$. Now all we need to do is show that $-\frac{\tan(x)}{\sqrt{x}}$ goes to 0 as $x \to 0$. It turns out that $\tan(x) < 2x$ for $0 < x < 1$, so $-\frac{\tan(x)}{\sqrt{x}} > -\frac{2x}{\sqrt {x}} = -2\sqrt{x}$, which goes to 0 as $x \to 0$.

So that's it done! We successfully sandwiched $\ln(x)\tan(x)$ between $0$ and $-2\sqrt{x}$, both of which go to 0 as $x \to 0$.

Hope that saves the day!

-- Uncle Colin

- discovered by Bernoulli, of course [↩]