How to show $f,g$ are equal up to n'th order at $a$. I need help with this problem.
Let two real-valued functions $f,g$ be equal to the $n'$th order at $a$ if 
$$\lim_{h\rightarrow 0}\frac{f(a+h)-g(a+h)}{h^n} = 0 $$
If $f'(a),\ldots,f^n(a)$ exist, show that $f$ and the function $g$ be given by
$$g(x) = \sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i$$
are equal up to the $n$'th order at $a$.
I'm also given the following hint.  The limit
$$\lim_{x\rightarrow a}\frac{f(x) -\sum_{i=0}^{n-1}\frac{f^{(i)}(a)}{i!}(x-a)^i}{(x-a)^n} $$
may be evaluated using L'Hospital's rule.
Can't I just plug in $g(a+h)$ and apply L'Hopital $n$ times?  I'm not sure what to do with this hint.
 A: If you plug $x-a$ into $h$ in the definition of $n$th order, you end up evaluating that limit.  You can then apply L'Hopital's Rule enough times to evaluate the limit.
UPDATE:
The sum only goes up to $n-1$.  That's because you can separate out the $n$th term to get
$$
-\frac{\frac{f^{(n)}(a)}{n!}(x-a)^n}{(x-a)^n}=-\frac{f^{(n)}(a)}{n!}
$$
for $x\neq a$.
A: The definition you are given is equivalent to 
$$\lim_{x \to a} \frac{f(x)-g(x)}{(x-a)^n}=0 $$
We are given $g$ defined as the sum
$$g(x) = \sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i$$
so we want to check wether or not 
$$\lim_{x \to a} \frac{f(x)-\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i}{(x-a)^n}=0 $$
Since the last term has $(x-a)^n$ in it, we can take it out and get
$$\lim_{x \to a} \frac{f(x)-\sum_{i=0}^{n-1} \frac{f^{(i)}(a)}{i!}(x-a)^i}{(x-a)^n}-\frac{f^{(n)}(a)}{n!}= $$
The sum is a polynomial in $(x-a)$ of degree $n-1$, which, as well as $f$, has its $k^{th}$ derivative defined for $k=0,1,\dots,n$, so we can use L'Hôpital Bernoulli to find the indeterminate limit. Applying it $n-1$ times gives
$$\frac 1 {n!}\mathop {\lim }\limits_{x \to a} \frac{{{f^{\left( {n - 1} \right)}}(x) - {f^{\left( {n - 1} \right)}}\left( a \right)}}{{\left( {x - a} \right)}} - \frac{{{f^{(n)}}(a)}}{{n!}} = $$
And by definition of the derivative you get
$$\frac{{{f^{(n)}}(a)}}{{n!}} - \frac{{{f^{(n)}}(a)}}{{n!}} = 0$$
Note we can apply L'Hôpital in every $n-1$ step because it is the case in every step we face a $\frac 0 0 $ indeterminate form.
What you are asked to prove is usually stated as
$$f(x) = T_n^a(x)+o((x-a)^n)\text{ ; when } x \to a$$
and means the error is of greater order (it goes to zero faster) than $(x-a)^n$ when $x \to a$. Though confusing, the meaning of "greater order" might be cleared up with this example: $(x-a)^{n+1}$ is of greater order than  $(x-a)^n$, and thus goes to $0$ "faster" than it for $x \to a$, since 
$$\mathop {\lim }\limits_{x \to a} \frac{{{{\left( {x - a} \right)}^{n + 1}}}}{{{{\left( {x - a} \right)}^n}}} = \mathop {\lim }\limits_{x \to a} \left( {x - a} \right) = 0$$
