You do it the same way we do with numbers: find a common denominator and then just add the numerators.
For example, if you needed to add
$$\frac{1}{2}+\frac{1}{3}$$
you need to find a "common denominator" (a number which is a multiple of both $2$ and $3$), convert the two fractions to the common denominator, and then add them. The simplest common denominator is just the product of the two denominators, we one way to do this is:
$$\begin{align*}
\frac{1}{2}+\frac{1}{3} &= \frac{1}{2}\cdot\frac{3}{3} + \frac{1}{3}\cdot\frac{2}{2}\\
&= \frac{3}{6} + \frac{2}{6}\\
&= \frac{3+2}{6}\\
&= \frac{5}{6}.\end{align*}$$
So do the same thing here:
$$\begin{align*}
\frac{4x+4h}{x+h+1} - \frac{4x}{x+1} &= \frac{4x+4h}{x+h+1}\cdot\frac{x+1}{x+1} + \frac{4x}{x+1}\cdot\frac{x+h+1}{x+h+1} \\
&= \frac{(4x+4h)(x+1)}{(x+h+1)(x+1)} - \frac{4x(x+h+1)}{(x+h+1)(x+1)}.\end{align*}$$
Now you can just add them, since they have the same denominator, by keeping the denominator and adding numerators. Do the algebra in the numerator, and simplify as needed.
Note: If you are more comfortable with the algebra, there are things you can do to simplify your life; when adding fractions, sometimes the product of the denominators is not the best common denominator: if you can spot a simpler one, use it. For instance, if we had $\frac{1}{6}+\frac{1}{10}$, then we could use $30$ as the common denominator, instead of $6\times 10 = 60$, which leads to smaller numbers, easier arithmetic; likewise, if we had
$$\frac{1}{x^2-1} + \frac{2}{x^2+2x+1}$$
then you could use $(x^2-1)(x^2+2x+1)$ as a common denominator, but if you could spot that $x^2-1 = (x+1)(x-1)$ and $x^2+2x+1 = (x+1)^2$, then you might realize that you can use $(x+1)^2(x-1)$ instead, which is smaller (degree 3 instead of degree 4).
In your situation, on possible simplification is to note that both fractions are multiples of $4$. So you can factor out $4$ and deal with simpler fractions if you want. But if you are unsure about these kinds of simplifying steps, then don't do them. They are not strictly necessary, though they are useful when dealing with more complicated expressions.
Firstly, $\frac{p}{0}$ never becomes $\infty$, it is simply undefined.
Now, a fraction $\frac{p}{q}$, with $p,q$ integers and $q\ne 0$, is introduced in order to solve a problem. The problem is to solve the equation $q\cdot x = p$. This solution, even though described with integers alone, does not always have a solution in the integers. So, we extend the integers by introducing solutions to all such equations as long as it makes sense. An example where it does not make sense is the equation $0\cdot x = p$, where $p\ne 0$, since we wish to retain the fact that $0\cdot x=0$. There are some subtleties here, solutions to different equation may lead to essentially that same fraction. This is solved by the familiar equations of the form $\frac{1}{2}=\frac{2}{4}$ and so on.
Now, fractions with negative denominators simply are solutions for equations of the form $-7\cdot x = 2$ or $-5x=17$. There is no reason to exclude such solutions. In fact, by adjoining all solutions to all equations of the form $qx=p$, where $p,q$ are integers and $q\ne 0$ (and properly identifying fractions that only appear to be different in form) one obtains the field of the rational numbers.
Best Answer
$\frac{50}{100},\frac{51}{101},\frac{52}{102}$
The numerator grows bigger in proportion to itself compared to the denominator. Meaning the numerator was multiplied by a bigger number than the denominator. Hence the fractions grow in size as we add more terms to the sequence.
However, in proportion of multiplication the factor reduces, as for example from $1.0099$ to $1.0096$ since adding one to bigger numbers becomes less significant.
Bonus: The sequence is approaching one. I can't put up the plot here but for this specific sequence plot $f(x)=\dfrac{x}{x+50}$ and see what $\displaystyle \lim_{x \to \infty} f(x)$ is.
In addition, if you check the slope of the graph either by calculus or looking, it becomes less steep for bigger values of $x$ which also kinda answers your question.