What does it mean to "have all (small) limits"? If a question were to ask me to show that a category has all limits, is this dependent on what the functor and index category are, or is it indepent of this? Are there tricks to showing that a category does indeed have all small limits? 
 A: "The category $\mathcal C$ has all small limits" means "for each $\mathcal D$ a small category, and each $F: \mathcal D \to\mathcal C$, the limit of $F$ exists". 
The trick, typically, is showing that a category has all equalizers and all small products; then the limit of a functor $F: \mathcal D \to \mathcal C$ is the equalizer of the two arrows
$$
\sigma,\tau :\prod_{D \in \mathsf{ob}(\mathcal D)} F(D) \to \prod_{f: D \to D' \in \mathsf{ar}(\mathcal D)}F(D')
$$
where $\sigma$ is defined by $\pi_{f: D \to D'} \circ \sigma = \pi_{D'}$, and $\tau$ by $\pi_{f: D \to D'} \circ \tau = F(f) \circ \pi_D$. If we call this equalizer $e: E \to \mathrm{dom}(\sigma)$, then $E$ becomes a limit of $F$ via the projections $\pi_D \circ e: E \to F(D)$. By construction, for $f: D \to D'$ an arrow of $\mathcal D$,
$$
\begin{align*}
F(f) \circ \pi_D \circ e &= \pi_{f:D \to D'} \circ \tau \circ e\\
& = \pi_{f: D \to D'} \circ \sigma \circ e\\
& = \pi_{D'} \circ e.
\end{align*}
$$
Conversely, any cone on $F$ must obey the above relations, and thus equalize $\sigma,\tau$, and thus factorize through $e$. (Fill in the details yourself.)
