Suppose $F : I \to \mathcal C$ is a diagram for a small category $I$ and a category $\mathcal C$. Suppose also that there is an object $L$ of $\mathcal C$ such that for all $X \in Obj(\mathcal C)$ there is an isomorphism $$ \alpha_X : \text{Mor}_{\mathcal C}(X, L) \simeq \displaystyle\lim_I \text{Mor}_{\mathcal C} (X, F(i)) $$ naturally in $X$. Prove that $\lim F$ exists, and that $L$ is isomorphic to $\lim F$.
I am new to category theory and I am having difficulties parsing through everything. We know that, since the category of sets have all limits and colimits, the given condition makes sense. I have proven that, IF there exists $\lim F$, then there exists such an isomorphism between the two sets for each $X \in Obj(\mathcal C)$. I am wondering how I can solve the above. My instructor suggested that I use Yoneda's lemma, along with the functor $G: \mathcal C^{op} \to \textbf{Set}$ with $G(X) = \lim \text{Mor}(X,F(i))$. However, this is where I am stumped: he seemed to suggest that $G(X)$ thus defined can be characterized as the set of functions $(\zeta_i: X \to F(i))_{i \in I}$ such that for all $\theta: i \to j$, we have that the diagram [ \begin{array}{ccc} X & \xrightarrow{\zeta_i} & F(i) \\ & \searrow_{\zeta_j} & \downarrow^{F(\theta)} \\ & & F(j) \end{array} ] commutes. However, I am not able to see that, since the coproduct $G(X) := \prod \text{Mor}(X,F(i))$ comes only with the obvious projection map to the sets $\text{Mor}(X,F(i))$. The instructor then noted that, by Yoneda's lemma, there is a unique natural transformation $\alpha : \text{Mor}(-,L) \to G$ that corresponds to $G(L)$, and said that the condition that $\alpha_X$ is bijective for all $X$ is satisfied exactly when the cone for $L$ is final.
All of this is very confusing to me. Also, why does the question mean when it says that the isomorphism holds "naturally in $X$"? Does it mean that we have a natural isomorphism between two functors?
$\newcommand{\morc}{\mathsf{Mor_C}}$Well, one issue is that this is a perfectly valid way of defining limit and the answer to the question depends on what definition you view as the fundamental one. There are quite a few candidates and I do not claim that that list is complete.
The intuition for this: that isomorphism just tells you that arrows into $L$ are the same thing as cones over $F$. This is just the statement (loosely speaking) that $L$ limits $F$.
(a variant of) Yoneda's lemma tells you that $\alpha$ can be encoded as a "nice" element $\beta$ of $\varprojlim_{i\in I}\morc(L,F(i))$. Specifically, the fact $\alpha$ is an isomorphism tells you that $(\beta,L)$ is initial in the category of elements of the functor $\varprojlim_{i\in I}\morc(-,F(i))$. If you don't know what that means, that's fine; in concrete terms, the isomorphism tells you that for any $X$ and $\gamma\in\varprojlim_{i\in I}\morc(X,F(i))$ there shall be some unique arrow $\delta:X\to L$ such that $(\varprojlim_{i\in I}\morc(-,F(i)))(\delta)(\beta)=\gamma$. $\beta=\alpha_L(1_L)$ and the corresponding $\delta$ shall be $\alpha_X^{-1}(\gamma)$. Take a moment to convince yourself this is really true.
I assume you use the definition of limit via universal cones. To unwrap what the above paragraph is saying into a proof about existence of limits we just need to see that $\varprojlim_{i\in I}\morc(-,F(i))$ is isomorphic to the functor of cones i.e. it maps $X$ to the set of all cones from $X$ to the diagram $F$ and acts on arrows $f:X\to Y$ by taking a cone $\varphi:\Delta Y\implies F$ to $\varphi\circ f$. Then the universality of $(\beta,L)$ tells you in fact that there is some universal cone $\hat{\beta}$ from $L$ to $F$, and the contravariance flips the initiality of $(\beta,L)$ into terminality of $(\hat{\beta},L)$ (! intuitively speaking... more precisely you can use the explicit property of $\beta$ that I mentioned to conclude) which is exactly what we need as per the definition of a limit.
I hope you know the explicit construction of limits in the category of sets (see bottom of this post). Using this, I see that (... for any fixed choice of limit functor $\varprojlim_{i\in i}$, I will get something isomorphic to the construction I describe): $$\varprojlim_{i\in I}\morc(X,F(i))\\\cong\left\{(\lambda_i)_{i\in I}\in\prod_{i\in I}\morc(X,F(i)):\forall(f:i\to j\text{ in $I$}),\,\morc(1,F(f))(\lambda_i)=\lambda_j\right\}$$Notice each $\lambda_\bullet$ is just an arrow $X\to F(\bullet)$ and the condition on the right is saying exactly that $F(f)\circ\lambda_i=\lambda_j$ always holds (when $f:i\to j$) so this is precisely the statement that the $(\lambda_i)_{i\in I}$ define a cone over $F$ with apex $X$.
Once you realise this, take another moment to convince yourself this correspondence of elements of $\varprojlim_{i\in I}\morc(X,F(i))$ with cones is natural in $X$ (how is $\varprojlim$ defined as a functor on arrows? What is the limiting cone in the explicit construction of limits in sets, and what does that say about how this construction behaves when it acts on some $g:X\to Y)$?
Reading your question again, it seems that maybe you're not happy with that formula after all. In general for a functor $G:\mathsf{C}\to\mathsf{Set}$, where $\mathsf{C}$ is small, $\varprojlim_{\varsigma\in\mathsf{C}}G$ can be constructed as: $$\left\{(x_\varsigma)_{\varsigma\in\mathsf{C}}\in\prod_{\varsigma\in\mathsf{C}}G(\varsigma):\forall(f:\varsigma\to\varsigma')\text{ in $\mathsf{C}$},\,G(f)(x_\varsigma)=x_{\varsigma'}\right\}$$You can justify this (I leave this as an excellent conceptual exercise) by thinking: well, if the limit really exists - call it $L$ - I know $L\cong\mathsf{Set}(\ast,L)$ where $\ast$ is some singleton set, and then I know arrows $\ast\to L$ should be the same thing as cones over $G$ with apex $\ast$... but arrows $\ast\to G(\varsigma)$, for any $\varsigma$, are just elements of $G(\varsigma)$. What does the cone condition say about these (families of) elements?