Three questions that I failed to approach properly (statements, true or false). I have these three questions: one about power series (true or false), and one about a statement about derivative and continuity (that I'm still not sure why it's true), and one about limits, that ruined my confidence in solving these questions and I'll explain my thought process for each one, I will appreciate any tips about how to approach these type of questions so I can improve myself. 
True or false: If  $\sum^\infty_{n=0}a_nx^n$  converges exactly at $[-2,2)$, then $\sum^\infty_{n=0}a_n(x^2+1)^n$ converges exactly at $(-1,1)$.
$\textbf {My Work}$: I said let $t=x^2+1$, then $\sum^\infty_{n=0}a_nt^n$ converges at $[-2,2)$, now $-2\le t < 2$, then $-3\le x^2 < 1$, Now I know that $x^2$ can never reach $-3$, for some reason I went with that and inferred that $x$ can be whatever in the negative side $(-\infty,0$], ( I guess I understand my mistake right now but I will put this here maybe I can get some tips about how I approached the question). 
Second question: If the right and left derivatives of $f(x)$ exist at $0$, then $f(x)$ is continuous at $0$. 
$\textbf {My Work}$:  I know that if the derivative exists at point, then that means the function is continuous at that point, but what went through my mind is this: If the right and left derivatives exist but they're not equal, that means $f(x)$ isn't differentiable at $x=0$, which means it's not surely continuous, what surprised me that this is true, still can't get why.
Third and last question about limits: let $f(x)$ be differentiable for all $x$, then which of these is true? 
a) if $\lim_{x\to\infty}\frac {f(x)}{x}=\infty $ ,then $\lim_{x\to\infty}f'(x)=\infty$. 
b) if $\lim_{x\to\infty}f(x)=1$ then $\lim_{x\to\infty}f'(x)=0$. 
c) if $\lim_{x\to\infty}f'(x)=8$ then $\lim_{x\to\infty}f(x)=\infty$.
$\textbf {My Work}$: 
a) I didn't know how to approach, it looked very true but it was false in the end. 
b) I couldn't find a counter example, but from intuition, from drawing a graph, a function can converge to a limit in a weird way (like it goes up and down fast), and the derivative limit might not exist, couldn't find a counter example (thought of functions with $sin$ or $cos$ but didn't succeed).
c) It's the answer, but I couldn't figure out why it's true rather than intuition which fails me most of the time sadly. It would mean alot to me to get tips about my thought process and how I should approach these questions, (especially the third one), any tip is really appreciated, thanks in advance to everyone.
 A: (1) We want to know which values of $x$ will make the second series converge. If $x\le1$ or $x\ge1$, then $x^2+1\ge2$, so we're plugging a value into the first series that's not in $[-2,2)$, so it diverges. On the other hand, if $x\in(-1,1)$, then $x^2+1\in[1,2)$, which is a subset of $[-2,2)$, so we're plugging in a value that makes the series converge. In other words, $(-1,1)$ is exactly the set of $x$ values for which $x^2+1\in[-2,2)$. This question isn't really about power series at all, just sets and functions.
(2) You're right that $f$ might not be differentiable at $0$, but it is necessarily continuous. The left and right derivatives both take into account the value $f(0)$ (recall the limit definition of the derivative), and for both derivatives to exist, $f(x)$ must approach $f(0)$ as $x$ approaches $0$ from either side. That's exactly what it means for $f$ to be continuous at $0$.
(3) Your idea for (b) is exactly right! A counterexample is $f(x)=1+\sin(e^x)/x$. For a similar reason, (a) is false: picture something like $f(x)=x^2+\sin(e^x)$. Your mental image for (c) is probably right too. Since $f'(x)$ converges to 8, the slope of $f$ at some point becomes larger than 7 and stays that way forever, so $f$ increases without bound.
