Consider the ideal I of Q[x] generated by the two polynomials f = x^2+1 and g=x^6+x^3+x+1
a) find h in Q[x] such that I=<h>
b) find two polynomials s, t in Q[x] such that h=sf+tg
Can someone, please, show me an example of when you are better of with parabolic cylindrical coordinates than with cartesian coordinates when computing a triple integral over a solid?
\cos \left( {2x} \right) = \cos ^2 x - \sin ^2 x = (cos x - sin x)(cos x + sin x)
does it ring a bell now? You have to do something with the numerator.
To prove that lim f(x,y) does not exist, it suffices to show that the limit along one curve into (a,b) differs from the limit along a second curve. If lim f(x,y) does exist, however, then computing limits along individual curves will prove nothing (although, such computations will likely help to...
You can show that the limit is undefined by showing that the limit depends on how you approach zero.
You can show that f(x,y) \longrightarrow \frac{1}{2} if you approach zero along the curve (\sqrt{y},y).
So you have showed that the limit is both 0 and 1/2, i.e. it must be undefined.
You seem to have misunderstood L'Hospitals rule.
L'Hospitals rule states that \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)} if \lim f(x) and \lim g(x) are both zero or ±\infty. L'Hospitals rule isn't necessary. Just use the standard limit for \frac{sin(x)}{x}