Ways to solve these types of indeterminations? I'm looking for this type of indeterminations and what I have to do to solve these:
$\lim\limits_{x\to\infty}{f(x)=\dfrac{\infty}{\infty}}$
$\lim\limits_{x\to\infty}{f(x)=\infty-\infty}$
$\lim\limits_{x\to\infty}{f(x)=\dfrac{0}{0}}$
$\lim\limits_{x\to\infty}{f(x)=\dfrac{n}{0}},n\not=0$

EDIT:
Could anyone post any examples of these indeterminations?


Could anyone post any examples of these indeterminations?

It's too long to evaluate in detail  indeterminations of all the 4 types, with and without using L'Hôpital's rule. Just one 
Example of $\infty -\infty $ indeterminations.
Sometimes these indeterminations can be evaluated without using l'Hôpital's rule. That's the case of rational fractions in $x$, i.e $\frac{P(x)}{Q(x)}$, with $P(x)$ and $Q(x)$ polynomials in $x$. As an example let $f(x)=\frac{x^{2}}{x+2}$ and $g(x)=\frac{x^{3}}{x+3}$. We have
$$\begin{eqnarray*}
\lim_{x\rightarrow \infty }f(x) &=&\lim_{x\rightarrow \infty }\frac{x^{2}}{x+2}=\infty  \\
\lim_{x\rightarrow \infty }g(x) &=&\lim_{x\rightarrow \infty }\frac{x^{3}}{x+3}=\infty. 
\end{eqnarray*}$$
Hence $\lim_{x\rightarrow \infty }f(x)-g(x)$ is indeterminate. Since we can rewrite $f(x)-g(x)$ as
$$\begin{equation*}
\frac{x^{2}}{x+2}-\frac{x^{3}}{x+3}=\frac{-x^{4}-x^{3}+3x^{2}}{x^{2}+5x+6}:=
\frac{P(x)}{Q(x)},
\end{equation*}$$
where $P(x)=-x^{4}-x^{3}+3x^{2}$ and $Q(x)=x^{2}+5x+6$, we have
$$
\begin{eqnarray*}
\lim_{x\rightarrow \infty }\frac{x^{2}}{x+2}-\frac{x^{3}}{x+3}
&=&\lim_{x\rightarrow \infty }\frac{P(x)}{Q(x)} \\
&=&\lim_{x\rightarrow \infty }\frac{-x^{4}-x^{3}+3x^{2}}{x^{2}+5x+6} \\
&=&\lim_{x\rightarrow \infty }\frac{-x^{2}-x+3}{1+5/x+6/x^{2}} \\
&=&\frac{\lim_{x\rightarrow \infty }-x^{2}-x+3}{\lim_{x\rightarrow \infty
}1+5/x+6/x^{2}} \\
&=&\frac{-\infty }{1+0+0}=-\infty.
\end{eqnarray*}$$
The polynomials $P(x)$ and $Q(x)$ are differentiable. We can thus apply l'Hôpital's rule to the fraction 
$$\begin{equation*}
\frac{P(x)}{Q(x)}=\frac{-x^{4}-x^{3}+3x^{2}}{x^{2}+5x+6}
\end{equation*}$$
as follows
$$\begin{eqnarray*}
\lim_{x\rightarrow \infty }\frac{P(x)}{Q(x)} &=&\lim_{x\rightarrow \infty }
\frac{P^{\prime }(x)}{Q^{\prime }(x)}\\
&=&\frac{\lim_{x\rightarrow \infty }-4x^{3}-3x^{2}+6x}{\lim_{x\rightarrow
\infty }2x+5} \\
&=&\frac{\lim_{x\rightarrow \infty }-12x^{2}-6x+6}{\lim_{x\rightarrow \infty
}2}=-\infty .
\end{eqnarray*}$$
The final results are, of course, the same. The evaluation of
$$\displaystyle \lim_{x \to 4} \; \frac{x-4}{5-\sqrt{x^2+9}}$$
is done in this question. There are many other examples in this site.
Exercise. Try to evaluate the similar indetermination in the limit of 
$$\begin{equation*}
\frac{1}{x-3}+\frac{5}{\left( x+2\right) \left( 3-x\right) }
\end{equation*}$$
as $x$ tends to $3$.
