Repeated application of L'Hôpital on polynomials doesn't always work? This is a follow up to a question posted yesterday here. 
For my question, suppose you wish to compute the following limit
$$
\lim _{x\to 0^+}\left(\frac{2x^7+x^2}{x^7+5x^4}\right) 
$$
My first thought was to apply L'Hôpital's rule 7 times because the numerator and denominator both tend to zero. Without doing any calculations, it should be obvious that if you take 7 derivatives of the top and bottom (i.e. apply L'Hôpital's rule)  you'll end up with a ratio of constants which I don't believe is the correct answer.  
After performing polynomial long division I reduced the limit to the following
$$
\lim _{x\to 0^+}\left(2 +\frac{x^2-10x^4}{x^7+5x^4}\right) =\lim _{x\to 0^+} 2 + \lim _{x\to 0^+}\left(\frac{x^2-10x^4}{x^7+5x^4}\right)
$$
Taking the derivative 4 times of the top and bottom of the second limit on the right yields the following
$$
\lim _{x\to 0^+}\left(\frac{-240}{840x^3+120}\right) 
$$
Hence we have 
$$
\lim _{x\to 0^+}\left(2 +\frac{x^2-10x^4}{x^7+5x^4}\right) = 2  - 2 = 0
$$
Which I still don't think is the correct answer. What's going on here!
Question: Why does the repeated naive application of L'Hôpital's rule appear not to be working
PS: According to Wolfram Alpha the answer should be $+\infty$. 
 A: L'Hospital is proven and cannot "not work". Never. Provided you stick to the theorem hypothesis, among which: the numerator and denominator are going to zero. This is where your reasoning fails.

Actually, this limit doesn't require L'Hospital, as
$$
\lim _{x\to 0^+}\frac{2x^7+x^2}{x^7+5x^4}
=\lim _{x\to 0^+}\frac{2x^5+1}{x^5+5x^2}$$
which is not an indeterminate form.
A: No, you don't apply L'Hôpital seven times in the original form of the function, that would be very wrong, just like
$$
\lim_{x\to0^+}\frac{x}{x^2}=\lim_{x\to0^+}\frac{1}{x}=\lim_{x\to0^+}\frac{0}{1}=0
$$
Can you spot where the wrong $=$ is? Because you certainly know the limit is $\infty$.
When you differentiate twice the numerator you get
$$
84x^5+2
$$
and this function has limit $2$ as $x\to0$, so it is not one for which L'Hôpital applies.
If you had the limit at $\infty$, then applying L'Hôpital seven times would be possible and would also give the correct result, because up to the sixth step you still have functions that have infinite limit at $\infty$. Possible, but boring and inefficient.
Be careful with L'Hôpital: a good tool, but dangerous if not handled with care.
A: You have to check hopital hypothesis after every differentiation. In particular after two differentiation the numerator won't be tending to $0$ anymore (there will be a constant term) so you have to stop there.
There also other problems, namely:
$$\lim _{x\to 0^+}\left(2 +\frac{x^2-10x^4}{x^7+5x^4}\right) =\lim _{x\to 0^+} 2 + \lim _{x\to 0^+}\left(\frac{x^2-10x^4}{x^7+5x^4}\right)$$
This is only valid if both limits on the RHS exist, something which you have to justify. 
Also, there really is no reason to use long polynomial division instead of just applying hopital directly
Finally, a cool way to do it is using infinetesimals; in fact since 
$$2x^7 + x^2 \sim_0 x^2$$ and $$x^7 + 5x^4 \sim_0 5x^4$$, we find that 
$$\frac{2x^7 + x^2 }{x^7 + 5x^4} \sim \frac{x^2}{5x^4} = \frac 1{5x^2} \to \infty$$
A: It is in fact $+\infty$ by the presence of $x^4$ in the denominator and $x^2$ in the numerator. After you apply twice the Rule you have $84x^5+2$ as numerator.
