Recent content by Rasmus

Thanks for the help, but I'm afraid I don't really understand exactly what you have done. It looks like you applied arcsin twice as well as some other operations to solve for \phi rather than r. Wouldn't arcsin be a problem since that isn't defined for [0 to 2pi]?

Homework Statement Write the equation x^2 + y^2 = 1 + sin^2(xy) in polar form assuming x = rcos(\phi) y = rsin(\phi) 0<r, 0<= \phi < 2pi solve for r as a function of \phi The Attempt at a Solution (rcos(\phi))^2 + (rsin(\phi))^2 = 1 + sin^2(r^2cos(\phi)sin(\phi))...

Thanks for the hint! As far as I understand the problem we start by 1/2 and multiply another rational for each increment increase in n. Meaning that a_{n+1} = a_{n} * (2n)/(2n +1). Since n is a natural number that menas the latter factor is less than one, therefore a_{n} must be greater than...

Homework Statement Show that a_{n} = \frac{1 \cdot 3 \cdot 5 \cdot ... \cdot (2n - 1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot 2n} converges when n \rightarrow∞ and n is a natural number Homework Equations None that I can think of.The Attempt at a Solution This was from an exam and I was pretty...

Here's my thinking: \frac{1}{e^{2x} - 1} \cdot ln(1 + sin^2 x) = ln(1 + sin^2 x)^{\frac{1}{e^{2x} -1}} When x approaches 0 ln(1 + sin^2 0)^{\frac{1}{e^{0} - 1}} = ln(1 + 0)^{\frac{1}{0}} = ln(1)^{∞} = ln(1) = 0 Correct?

Edit: Nevermind I misunderstood I don't really understand how that expression evaluates though.

Not a clue :) I did however get help with a solution elsewhere. \lim_{x\to 0} \frac{ln(1 + sin^2x)}{e^{2x} - 1} First multiply by \frac{sin^2x}{sin^2x} Then \frac{ln(1 + sin^2x)}{sin^2x} is on standard form and will approach one as x approaches zero. \frac{sin^2x}{e^{2x}} is...

Homework Statement Find the limit (without using L'Hospital's rule) \lim_{x\to 0} \frac{ln(1 + sin^2x)}{e^{2x} - 1} Homework Equations The Attempt at a Solution I tried various substitutions in order to rewrite the expression to a standard limit. Such as t = e^x, t = ln(x), t =...