The above answer does it totally. However you can visualize this problem graphically. Imagine the three axes $x,y,z$ and the first quadrant part of the curve $xy=1$ drawn on the $xy$ plane. The ...

Yes it is continuous. Compositions of two continuous functions is always continuous. In this case you can see it by the sequential definition of continuity. $$x_n\rightarrow x \Rightarrow f(x_n)\... View answer 2 votes You are right. Use the fact$$\lim_{x\rightarrow\infty}\bigg(1+\frac{1}{x}\bigg)^x = e$$View answer 1 votes I am not sure whether you mean a) and b) in general or just in a discrete metric space so let me answer for both cases A) for the discrete metric space it is always true by definition. for a general ... View answer 1 votes since it is analytic so the complex derivative computed along x-direction and y-direction is the same and is the same as the derivative. you compute the derivative in the x-direction. it is real. so ... View answer Accepted answer 1 votes Join the two centers and extend the line. wherever it cuts the two circles at their outer extremes are the points which are crucial say A and B. take the mid point of these two points. for ... View answer Accepted answer 1 votes You are looking for a linear (or affine) isometry which maps q_1 to t_1 and q_2 to t_2. First map q_1 to t_1. This gives a translation x\mapsto x + t_1 - q_1. call this map T. Then ... View answer 0 votes Go by the meaning of the derivative. It means when you change the parameters by a small amount, the value changes linearly as a function of this change. ie,$$f(x+\Delta x) \approx f(x) + df(x).x ...

PART - 2 I keep the notation as above. To wit, $P,Q$ are the fixed points on $L_1,L_2$. The vectors $v,w$ are the direction vector along $L_1,L_2$ and $P',Q'$ are the points to be found. Define two ...

PART 1 You probably want an algorithm to find the answer. I am ignoring the trivial case of parallel lines. Consider the points $P'$ and $Q'$ as moving points along their respective lines. Now ...
Note that you can write $\mathcal{P}(X)$ as set of all functions from $X$ to $\{0,1\}$ which I use as a standard 2 element set. So let me write $\mathcal{P}(X)$ as $\textrm{Map}(X,\{0,1\})$ If $Y$ ...