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What is the main reason you don't like it?

Let eps > 0. Take delta = min(2, eps/4). Let 0 < |x - 1| < delta. Then |x-1| < eps/4. Also, |x-1|<2, so -1 < x < 3, whence 0 < x+1 < 4. So |x+1| < 4. Factorise the x^2-1 into (x+1)(x-1). So |x^2 -1| = |(x+1)(x-1)| = |x+1||x-1| < 4*(eps/4) = eps. Thus the limit in question is 1.

Yeah.

For Q3, a direction vector for the line is the vector that the lambda is multiplying in your parametric form (that's the direction the line points in).

$\noindent Note $\tan \left(180^\circ - \theta \right) = - \tan \theta = -t$. So taking reciprocals, assuming $t\neq 0$, we have $\cot \left(180^\circ - \theta \right) =-\frac{1}{t}$.$

Re: MATH1131 help thread Because of the square root, the denominator is by definition positive (it can't be 0 since we can't have 0 on a denominator; and remember, the symbol √(t) refers to the positive positive square root of t, where t is a positive real number). Since y is equal to...

Re: MATH1131 help thread $\noindent The hyperbola thing you're thinking of is $y=\frac{1}{3-x}$. But this one is $y= \frac{1}{\sqrt{3-x}}$, which is different. Because of the square root in the denominator, we must always have $y>0$. Furthermore, it is easy to show that for any positive...

$\noindent This is assuming that $a\tan \theta$ is positive, of course. The general answer is $|a\csc \theta \sec \theta|$, assuming $\sec{\theta}\neq \pm 1$ (if it is $\pm 1$, then the given expression is undefined, as it has a zero denominator).$

Yes it does scale very well. It also makes your 3U Maths count for two units, so if you do well in 3U, it counts for two units of goodness rather than one unit of goodness. All in all, it probably is worth it for scaling, assuming you don't think it will obscenely detract from your time...

If it's from a 2U Question, you won't be expected to simplify it to sin(2α), since the double angle identities aren't part of 2U. So in that case you can leave it as 2*sin(α)*cos(α).

The double angle formula allows the 2*sin(α)*cos(α) to become sin(2α).

No, that's a fully simplified form.

If we can assume the extreme value theorem, we can do it as follows. Let f(\xi)=\Xi >0. By definition of limits to +/- infinity, there exist real numbers M>N such that |f(x)-0|<\Xi whenever x>M and whenever x<N. So \xi \in [N,M] (because outside this interval, $$f(x)\neq \Xi$$), and: (1)...

Re: HSC 2016 2U Marathon Yeah, with the 11 different sums, some of them can happen in more ways than others. So in actuality, there are 36 possibilities for the hand with two dice (Die 1 has 6 possibilities, and Die 2 has 6 possibilities).

Re: HSC 2016 2U Marathon I don't think it's that either.

Re: HSC 2016 2U Marathon I don't think it's 4/33.

The inverse is simply the function g: R -> R given by g(x) = cube root of x, where by cube root we mean the real cube root.

Yes.

The extreme value theorem applies to closed and bounded intervals only. The interval in your question is not closed, so the extreme value theorem doesn't apply here. You can see what's going on by sketching the graphs of those functions on the given interval.

Re: HSC 2016 3U Marathon $\noindent Let $n$ be a positive integer. Using mathematical induction or otherwise, show that $\frac{\mathrm{d} ^n}{\mathrm{d} x^{n}} \left(x^n\right) = n!$ (i.e. the $n$-th derivative is $n!$).$