About $\lim_{x\to\infty} f(x+1)/f(x)$ Consider $\lim\limits_{x \to +\infty} \frac{f(x+1)}{f(x)}$
Intuitively, it seems to me that this limit should equal to 1 because "infinity" and "infinity+1" is essentially the same thing. 
However, I'm not sure if this is completely correct.
Secondly, if it is correct, I have no idea how  to show this using the limit laws I have learned (or even where to begin since I don't know that either limit exists seperately so I can't just seperate them)
 A: Even under the assumption that $\lim_{x\to\infty} f(x) = \infty$ (implicit in your question), we cannot say much.


*

*The limit can be $1$:
take $f(x) = x$: $$\lim_{x\to\infty} \frac{x+1}{x} = 1$$

*The limit can be any positive number $a>1$:
take $f(x) = a^x$: $$\lim_{x\to\infty} \frac{a^{x+1}}{a^x} = a$$

*The limit can be $\infty$:
take $f(x) = e^{e^x}$: $$\lim_{x\to\infty} \frac{e^{e^{x+1}}}{e^{e^x}} = \infty$$

*The limit can fail to exist:
take $f(x) = \begin{cases} x &\text{ if } x\in 2\mathbb{N}\\2x &\text{ if } x\in 2\mathbb{N}+1\\ x &\text{otherwise.}\\\end{cases}$: then$$\lim_{n\to\infty }\frac{f(2n+1)}{f(2n)} = 2,\qquad \lim_{n\to\infty }\frac{f(n\pi+1)}{f(n\pi)} = 1$$
(to have the same "limit of the ratio fails to exist" behavior for a continuous function $f$, one can also consider $f(x) = x(2+\cos x)$: $\liminf_{x\to\infty} \frac{f(x+1)}{f(x)} = \frac{1}{2}$, $\limsup_{x\to\infty} \frac{f(x+1)}{f(x)} \simeq 1.73 > 1$, and indeed $\lim_{x\to\infty} f(x) = \infty$.)
Note: the ratio can have a limit $1\leq \ell\leq +\infty$, or no limit; in the latter case, it can even have a $\liminf$ equal to $0$. However, it cannot have a limit $\ell < 1$, as otherwise the function $f$ must be decreasing for $x$ big enough and then cannot satisfy $\lim_{\infty} f = \infty$.
A: Let consider as counterexample $f(x)=e^x$ then
$$\frac{f(x+1)}{f(x)}=\frac{e^{x+1}}{e^x}=e$$
or also 
$$f(x)=\sin (x)\implies \frac{f(x+1)}{f(x)}=\frac{\sin(x+1)}{\sin x}\to N.E.$$
Thus in general the limit depends upon the particular function considered.
A: Consider a random series taking the valus of $1$ or $2$
  with same probability: $r:\,\mathbb{N}^{+}\to\left\{ 1,2\right\}$ 
  where $\lim_{N\to\infty}\frac{1}{N}\sum_{i=1}^{N}r\left(i\right)=\frac{3}{2}$. The possible quotients $\frac{r\left(i\right)}{r\left(i+1\right)}$
  then take random values in $\left\{ \frac{1}{2},1,2\right\}$ 
  with probabilities $\frac{1}{4}$, $\frac{1}{2}$ and $\frac{1}{4}$
  respectively. There is no limit at all!
If however $\lim_{x\to\infty}f\left(x\right)$
  exists, what can we say about the quotient then?
Consider the series $h:\,\mathbb{N}^{+}\to\mathbb{Q}$
  with $h\left(i\right)=\frac{1}{2}^{i}$. Then $\frac{h\left(i\right)}{h\left(i+1\right)}=2$.
So we have a limit, but it is not $1$...
But there need not be a limit of the quotient even in the case that $f\left(x\right)$
  approaches a limit:
With our random function $r$
  as defined above, define a series $d:\,\mathbb{N}^{+}\to\mathbb{Q}$
  with $d\left(1\right)\,:=\,1$ and $d\left(i+1\right)=\frac{d\left(i\right)}{r\left(i\right)}$. The limit is $\lim_{i\to\infty}d\left(i\right)=0$. But $\frac{d\left(i\right)}{d\left(i+1\right)}=r\left(i\right)$
  takes no limit.
By continuos extension of the above series it is obvious there are similar examples for continuos functions. Use splines to construct examples for differentiable functions.
Bottomline: Without some additional knowledge about the function, it cannot even be guaranteed that the quotient takes any limit, let alone $1$.
