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We need to ask ourselves what the symbol √(z) actually means for a complex number z. In general, it is not true that √(zw) = √(z)√(w) for z, w complex. Recall that sqrt(·) is a multivalued function on the complex plane. E.g. sqrt(1) = +1 or -1. Now, we assign a principal value to this in the...

An easier way to see that N works is that for any 0 < eps < 1, we have 1/eps > (1/eps – 1) > 0, so (1/eps^2) > (1/eps – 1)^2, that is, N > M. Since we know that M works (by 'works', I mean that x being greater than it will imply that |f(x) – L| < eps), taking any larger value (e.g. N) will also...

The answer to b) is a correct one. The answer in c) is a correct one too. To see this, let eps > 0, N = 1/eps^2, and suppose x > N. Then x > 1/eps^2, which implies sqrt(x) > 1/eps. This implies that sqrt(x) > (1/eps) – 1, and this is equivalent to 1/(1 + sqrt(x)) < eps, i.e. |1/(1+sqrt(x)) –...

$\noindent Yes. The substitution of $u=x^{\frac{1}{6}}$ $\big{(}$so $u\to 2$, $x^{\frac{1}{3}}=u^2$ and $x^{\frac{1}{2}}=u^3$$\big{)}$ transforms the limit to $\lim _{u \to 2}\frac{u^2 - 4}{u^3 - 8}$, which can make it easier to see what to do.$

Try using L'Hôpital's rule or a substitution.

$\noindent The other way to do that Q. would be to just do it in cases as $x\to 0^{+}$ and $x\to 0^{-}$ (so that you can get the inequality signs correct etc.). As Drsoccerball said, you'll end up with both limits being 0. Then, since the limit from the left and the limit from the right will be...

$\noindent Yes. For example, just note that $0\leq \sin \frac{1}{x}\leq \frac{1}{x}$ for all $x\in \left [\frac{2}{\pi},\infty \right)$ $\Big{(}$this is seen by essentially applying the map $x\mapsto \frac{1}{x}$ to our earlier interval and method$\Big{)}$, and then use the squeeze law as $x\to...

$\noindent The above result is what allows us to go from $-|x|\leq |x|\sin \frac{1}{x}\leq |x|$ $\Big{(}$obtained by multiplying $-1\leq \sin \frac{1}{x}\leq 1$ through by $|x|>0\Big{)}$ to $-|x|\leq x\sin \frac{1}{x} \leq |x|$ (because the middle expressions in each inequality have the same...

$\noindent We shall prove the following general result, which is what justifies it (it's not hard to see is true or prove). Let $a> 0$ and $u,v \in \mathbb{R}$. Suppose $-a \leq u \leq a$ and $|u| = |v|$. Then $-a\leq v\leq a$.$ $\noindent \emph{Proof.} Note $-a \leq u \leq a$ is equivalent to...

$\noindent We can do it like this to make sure the inequalities are correct: $-|x|\leq x\sin \frac{1}{x}\leq |x|$.$

$\noindent It's just equivalent to $\lim _{\theta \to 0^{+}} \sin \theta$, which is 0. If you really wanted to use the pinching theorem, you could use the fact that for $\theta \in \left[0,\frac{\pi}{2}\right]$ say, we have $0 \leq \sin \theta \leq \theta$. Then by the pinching theorem (aka...

$\noindent You need to find each limit separately (using the behaviour of the function on the relevant side of $a$). Even if $f(a)$ is defined, there is no need for the limits to equal $f(a)$. And $f(-a)$ is unrelated to these limits.$

$\noindent Note that $x\neq 2$. Now, if $x>2$, we have $\frac{x-3}{x-2}=4,x>2$. Equivalently, $x-3 = 4x - 8 \Longleftrightarrow 3x = 5 \Longleftrightarrow x = \frac{5}{3}$. But this is not greater than $2$. So we have no solutions for $x>2$.$ $\noindent Now if $x<2$, we have...

Yes, you should state that, because the answer isn't fully simplified if left as -sin(x)/|sin(x)|.

That's because those are the domains where sin(x) is positive and negative, respectively.

For the first one, you obtained -sinx/|sin(x)|. This is just equal to the negative of the sign of sin(x). So when sin(x) is positive, -sinx/|sin(x)| = -1, and when sin(x) is negative, -sinx/|sin(x)| = 1. When sin(x) = 0, -sinx/|sin(x)| is undefined.

No, it's not wrong; you can use any point that lies on the plane, so there are (infinitely) many choices for it. The parametric equation is not unique.

$\noindent But yes, if there are just \emph{two} direction vectors in the parametric equation and they are non-zero and non-parallel, it will be a plane (and conversely, any plane can be written in such a parametric form).$

Well no. For example, if there were three vectors there and they were what is called linearly independent (which is equivalent to (although not defined as) saying that none of them is in the span of the other two ), then they would end up spanning the whole of ℝ3, rather than just a plane.

$\noindent If it was something like this: $\vec{x} = t_1 (1,0,0) + t_2 (2,0,0)$, then it's just a \emph{line} through the origin $\big{(}$in $\mathbb{R}^3$ of course$\big{)}$, because the two vectors are just scalar multiples of each other. Something like $\vec{x} = t_1 (1,0,0) + t_2 (2,1,0)$...