Proving the continuity of these maps Backstory: I am having an exam soon, and these are the assignments that keep coming up, I cannot finish any of them to the end, but have ideas about solving them, and would like to hear your thoughts on that, and if anyone can point me in the right direction in solving them. (Best case scenario is they get solved, but that would be unrealistic to expect). Anyway, they are based on what I believe is generally same idea.
1.Prove the continuity of the following functions:
a.)
$$f(x,y)= \begin{cases} \tan(x^3+y^3)\frac{\ln(x^2+y^2)}{x^2+y^2},(x,y)\neq(0,0) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1, (x,y)= (0,0) \end{cases}$$
b.)
$$f(x,y,z)=\begin{cases} (x^4+y^4)\frac{\sin(x^2+y^2+z^2)}{x^2+y^2+z^2}, (x,y,z) \neq(0,0,0) \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0, (x,y,z)=(0,0,0) \end{cases}$$
-These, first two I have no idea how to finish,or get very far at all, assignments that look similar in class had us choose $x_n \to 0$ and $f(x_n) $ wouldn't match up with the definition.
c.)
$$h(x,y)=\begin{cases}y-\frac{\sin x}{x}, x \neq0 \\ y-1,x=0 \end{cases}$$
-Here, this seems trivial, but am not sure if I am right: $\frac{\sin x}{x}\to^{x\to 0} 1.$ So the function is continuous in 0, and aswell as beyond zero, as a composition of continuous functions.
d.)
$$p(x,y)=\begin{cases}y- \frac{e^{x^2+y^2}-x^2-y^2}{x^2+y^2}, x^2+y^2 \neq 0\\ 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1,\  x=y=0 \end{cases}$$
-With this one it looks like to me it would be best to incorporate polar coordinates. But am not if that's the way to go, or how that would concretely help me.
e.)(This one interests me most)
$$f(x,y)= \begin{cases} \frac{\tan x}{x}+y,\  0 < \| x \| <1. \\ \ \ \ \  \ \ 1+y ,\ \  x=1 \end{cases}$$ on $(-1,1)\times \mathbb{R}$
2.Do the following functions map $x^2+y^2\leq1$ to a closed set ?
a.)
$$f(x,y)=\begin{cases} a-by\frac{e^x-1}{x}, x\neq 0 \\ \ \ \ \ \ \ \  a-by, x=0 \end{cases}$$ Here I would think proving continuity because $\frac{e^x-1}{x}\to 1 $ when $x\to 0$
b.)
$$f(x,y)=\begin{cases}a + \frac{b\tan(x^2+y^2)}{x^2+y^2}, (x,y)\neq (0,0)\\ a+b ,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \  (x,y)=0  \end{cases}$$
 A: 1) a. 
$$f(x,x) = \frac{\tan (2x^3)\ln (2x^2)}{2x^2} = \frac{\tan (2x^3)}{2x^3}x\ln (2x^2)$$ $$= \frac{\tan (2x^3)}{2x^3}x(\ln 2 + 2\ln |x|).$$
Now $\lim_{u\to 0}\tan u/u = 1,$ and $\lim _{x\to 0}x\ln (|x|)= 0.$ It follows that $\lim _{x\to 0}f(x,x) = 0\ne f(0,0).$ So $f$ is not continuous at $(0,0).$
(It's possible the problem was misstated?)
b. The function $g(u) = \sin u/u, u\ne 0, g(0)=1$ is continuous (in fact $C^\infty$) on $\mathbb {R}.$ Thus $g(x^2+y^2)$ is continuous on $\mathbb {R}^2, $ being the composition of continuous functions. Now $f(x,y) = (x^4+y^4)g(x^2+y^2)$ is the product of two continuous functions, hence $f$ is continuous.
c. Again look at the $g$ from b. above. We have $f(x,y) = y-g(x),$ a difference of two continuous functions, hence $f$ is continuous.
d. As $(x,y) \to (0,0), e^{x^2+y^2} \to 1.$ Thus $p(x,y)$ looks like $0 - (1-0)/0^+$ for $(x,y)$ close to $(0,0).$  Hence $p(x,y) \to -\infty$ as $(x,y) \to (0,0)$ and so is not continuous there. (Problem misstated?)
e. This is the same as b. and c. except this time we have $g(u) = \tan u/u, u\ne 0, g(0)=1,$  which is continuous on $(-\pi/2,\pi/2).$ So $f(x,y) = g(x) +y$ is continuous on $(-1,1)\times \mathbb {R}.$ (I assume you meant $f(x,y) = 1+y, x = 0.$)
2) a. We are doing the same thing here. Define $g(u) = (e^u-1)/u, u\ne 0, g(0)=1.$ Then $g$ is continuous on $\mathbb {R}.$ We have $f(x,y) = a-byg(x)$ everywhere, so $f$ is continuous on $\mathbb {R}^2.$ Now a continuous function maps a compact set to a compact set, so $f(\{x^2+y^2\le 1\})$ is compact, hence closed.
b. Totally the same. Use the $g$ from 1.e. We then have $f(x,y) = a + bg(x^2+y^2),$ hence $f$ is continuous. So $f(\{x^2+y^2\le 1\})$ is compact as above.
A: For 1,a. For $(x,y)\not=(0,0)$, let $M=max(abs(x),abs(y))$. Observe that $tan(x^3+y^3)=K(x,y)(x^3+y^3)$ where K(x,y) tends to $1$ as $(x,y)$ tends to $0$, and the absolute value of $x^3+y^3$ cannot exceed $2M^3$. Also $ abs(log(x^2+y^2))$ cannot exceed $log 2M^2 = (log 2)+2log M$, while the denominator $x^2+y^2$ is at least $M^2$. Finally, $MlogM$ goes to $0$ as M does.
