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 $3$.