TD

2sin(x)cos(x) = sin(2x) => sin(2x)' = 2cos(2x) 4cos²(x)-2 = 2(2cos²x-1) = 2cos(2x). It's all good

[math]\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b \Rightarrow \cos \left( {a + a} \right) = \cos \left( {2a} \right) = \cos ^2 a - \sin ^2 a[/math]

edit: too late.

You kind of solved the problem yourself. Vertical asymptotes will always appear when you have a fraction and certain values of x for which the denominator becomes 0 while the nominator doesn't. So a (simple) function of the form y = 1/(x-a) will always yield a vertical asymptote at x = a. Just substitute a to get a vertical asymptote at any desired x-value.

That is correct, a limit of a product may be calculated as a product of limits.

Eh, I misread: it says a k there, my bad Mathematica still says it doesn't converge though.

I don't think it exists when x = 0, since you then get 1^(1/0). Mathematica tells me it doesn't converge on [-1,1]...

It doesn't since the upper limit isn't equal to the lower limit (+ / - infinity).

1 is correct, your answer for 2 is correct (slope is 0 at that point, it has a very nice graph btw) and 3 doesn't seem right to me, but it's not really clear what you did.

That factor shouldn't be there I believe.

What you're saying isn't possible. You can let the last digit be recurring' date=' but not another one. So we're talking about 0.99[u']9[/u] and not 0.999, not that it really matters in this case. It does matter with 0.001, which isn't possible. You cannot have an infinite number of recurring 0's, and then still 'add a 1'.

Are you looking for a proof for the fact that you can get any limit by rearranging (in some cases, of course) or that you're allowed to rearrange when dealing with absolute convergent series?

Too bad, it's important ;o)

Thanks to the fundamental theorem of calculus, you can compute these (definite) integrals by using anti-derivatives. If F(x) is an anti-derivative of f(x), then: [math]\int\limits_a^b {f\left( x \right)dx} = \left[ {F\left( x \right)} \right]_a^b = F\left( b \right) - F\left( a \right)[/math] The middle step is just a notation and is not required. Can you figure it out now for that example?

Check http://www.math2.org/math/derivatives/more/trig.htm

Your algebra is correct, even according to my plot. Perhaps you made a wrong graph?

Well don't kick yourself, but do keep asking for help when you're stuck Good luck!

No problem, but as you can see it's not always necessary. For rather 'simple' problems, playing with powers will do

Use the fact that 9 = 3^2 and that (a^b)^c = a^(cb): [math]3^x = 9^{\left( {y - 1} \right)} \Leftrightarrow 3^x = \left( {3^2 } \right)^{\left( {y - 1} \right)} \Leftrightarrow 3^x = 3^{2 \cdot } ^{\left( {y - 1} \right)} \Leftrightarrow x = 2\left( {y - 1} \right)[/math] Or with logarithms: [math]3^x = 9^{\left( {y - 1} \right)} \Leftrightarrow \log _3 3^x = \log _3 9^{\left( {y - 1} \right)} \Leftrightarrow x = \log _3 3^2 ^{\left( {y - 1} \right)} \Leftrightarrow x = 2\left( {y - 1} \right)[/math]

By defnition (or equivalent with possible other definitions), we have that [math]\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{1} {x}} \right)^x = e[/math] Furthermore, we also have that in general, with m a real number [math]\mathop {\lim }\limits_{x \to \infty } \left( {1 + \frac{m} {x}} \right)^x = e^m [/math] So in your case, the answer would be [math]e^2 [/math]. To prove this, you can take the natural logarithm (ln/log) of the expression, the limit will then give the exponent of e.

Well normally you'd split by doing partial fractions but what I did doesn't require any advanced calculus. I added and substracted same things in the nominator (e.g. replaced "1" by "1+x-x" in the first step) and then split the fraction to simplify etc...

Yes, you can split using partial fractions or by "playing with fractions" like this:

No problem

Rewrite the fractions, expand the second one, add them and simplify. [math] \begin{gathered} \frac{x} {{2x - 2}} - \frac{1} {{x^2 - 1}} \hfill \\ \frac{x} {{2\left( {x - 1} \right)}} - \frac{1} {{\left( {x - 1} \right)\left( {x + 1} \right)}} \hfill \\ \frac{x} {{2\left( {x - 1} \right)}} - \left( {\frac{1} {{2\left( {x - 1} \right)}} - \frac{1} {{2\left( {x + 1} \right)}}} \right) \hfill \\ \frac{{x + 2}} {{2\left( {x + 1} \right)}} \hfill \\ \end{gathered} [/math] Now fill in x = 1