Proving the exponent of an element of a the Lie Algebra is an element of the Lie group. I have $X\in \mathfrak{sl}(n,\mathbb{R}) =\{X: \text{tr} (X) = 0\}$. And I need to prove that $A:=\exp(X) \in SL(n,\mathbb{R})=\{A: \det(A)=1\}$.

Let $A(t):=\exp(tX)$. 
Then we have that 
$$[X,Y]=0\implies\exp(X+Y)=\exp(X)\exp(Y)$$ 
we note that
$$[t_1X,t_2X]= t_1t_2X^2 - t_2t_1X^2 =0 $$ 
so we have
$$A(t_1 + t_2) = \exp(t_1X + t_2X)=\exp(t_1X)\exp(t_2 X)= A(t_1)A(t_2)$$
Let $f(t) := \det(A(t))$. 
Then 
$$f(t_1 + t_2) = \det(A(t_1+ t_2))=\det(A(t_1))\det(A(t_2))=f(t_1)f(t_2)$$
Finally,
$$\det(\exp(X)) = f(1) = f\left(k \cdot \frac{1}{k} \right) = \lim_{k\to\infty}  f\left(k \cdot \frac{1}{k} \right) 
= \lim_{k\to\infty} \prod_{n=1}^k f\left(\frac{1}{k} \right)
= \prod_{n=1}^\infty f(0) =\prod_{n=1}^\infty 1 = 1 
$$
where $f(0) = \det(exp(\mathbf{0}))=\det(\mathbf{1})=1$.
Is this proof correct? I'm particularly unsure about the limits. 
EDIT: the book I'm following says the answer should follow after "solving for $f(t)$" but I don't see how
 A: No, this is wrong and cannot be turned into a correct argument without major new ideas.  It is not correct to say that $$\lim_{k\to\infty} \prod_{n=1}^k f\left(\frac{1}{k} \right)
= \prod_{n=1}^\infty f(0).$$  For instance, if $f$ were the function $\mathbb{R}\to\mathbb{R}$ given by $f(x)=2^x$, then $\prod_{n=1}^k f\left(\frac{1}{k} \right)$ would be $2$ for all $k$ but $\prod_{n=1}^\infty f(0)$ would be $1$.
A different approach you might try is making a change of basis (over $\mathbb{C}$) to assume that $X$ is in Jordan normal form.  Then the trace and determinant depend only on the diagonal entries, and the diagonal entries of $\exp(X)$ can be computed by just exponentiating the diagonal entries of $X$.
A: I'm not sure how to justify the end of your argument, but a simpler way to prove that $\exp(X)\in \mathrm{SL}(n)$ is to use the identity $$\det\exp(A) = \exp(\textrm{tr} A)$$
that holds for all $n\times n$ matrices $A$. This identity is true because it is trivially true for all diagonalizable matrices and the subset of all diagonalizable matrices is dense on the space of all $n \times n$ matrices $\mathbb R^{n\times n}$, by a argument using the Jordan canonical form.
