The hard limits thread

In a metric space $(X,\delta)$, $\text{dist}\,(A,B)=\inf\{ \delta(a,b)\,|\,a\in A,\,b\in B\}$ where $A,B\subseteq X$. In vague terms it is the (closest) distance between two sets (note that this being zero does not imply that both sets share a common point).

The problem is not parallel to approximating reals with rationals: that is talking about $\mathbb{Q}$ being dense in $\mathbb{R}$. Here $\mathbb{Z}^*$ and $\alpha \mathbb{Z}^*$ are not subsets of each other, if $\alpha$ irrational, and one cannot pick an element of one and find elements in the other that are arbitrarily close to it. It is more a case two sets flirting with each other.

Right. It'd be nice to see some working though.

I'll have to come back to this after I've had some time to read up on metric spaces cos I've forgotten most of what I knew - you've confused me a bit with the statement "Here $\mathbb{Z}^*$ and $\alpha \mathbb{Z}^*$ are not subsets of each other" - isn't a set a subset of its closure? Maybe I'm missing the point.

Find $\displaystyle \lim_{x \to \frac{(2n+1)\pi}{2}} \text{cosec}^{\tan^2 x} x$

Find $\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{4n^2-k^2}}$

Find

Find $\displaystyle \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{\sqrt{4n^2-k^2}}$

Anyone know where I can find tricky problems without the need of appealing to Mr L'Hopital's Rule

...

then you can always make them up yourself with conditions that don't satisfy L'Hop's criterion.

I thought that this was very neat:

I don't know that this would be classified as 'hard', but it's a nice one:

$\displaystyle \lim_{x \to 0} \dfrac{\sin ax}{\sin bx}$

Give result with a rigorous proof .

Ah, this is nice. By L'Hop: $\displaystyle \lim_{x \to 0} \frac{\sin ax}{\sin bx} = \lim_{x \to 0}\frac{a\cos ax}{b\cos ax} = \frac{a}{b}$

Alternatively, Taylor expand: $\displaystyle \lim_{x\to 0} \frac{\sin ax}{\sin bx} = \lim_{x \to 0} \frac{ax + \mathcal{O}(x^3)}{bx + \mathcal{o}(x^3)} = \lim_{x \to 0}\frac{a + \mathcal{O}(x^2)}{b + \mathcal{O}(x^2)} = \frac{a}{b}$