Limit Problem - No clue where to start $$\lim_{x\to \infty}\frac{x^{2011}+2010^x}{-x^{2010}+2011^x}$$
I'm not sure where to even start with this one. One idea I had was that perhaps it could be "split" up as:
$$\lim_{x\to \infty}\frac{x^{2011}}{-x^{2010}+2011^x} + \lim_{x\to \infty}\frac{2010^x}{-x^{2010}+2011^x}$$
But solving this requires knowledge of the answer to the following question: What are the dominant terms? is $\infty^c > c^\infty$? It might even be an indeterminate form to which L'hopital's rule can be applied, but even if it was I'm not sure how to take the derivative of something to the power of x. D:
 A: We could begin by evaluating, for $\;x>0\;$ 
$$x^{2010}<2011^x\iff2010\log x<x\log2011$$
and the second inequality is clear for $
\;x\;$ big enough, so exponential rules over polynomial, and thus
$$\frac{x^{2010}}{-x^{2010}+2011^x}\xrightarrow[x\to\infty]{}0\\{}$$
$$\frac{2010^x}{-x^{2010}+2011^x}\xrightarrow[x\to\infty]{}0$$
A: Here's one approach: divide the top and bottom by $2011^x$ to get
$$
\lim_{x\to \infty}\frac{x^{2011}+2010^x}{-x^{2010}+2011^x} 
= 
\lim_{x\to \infty}\frac{\frac{x^{2011}}{2011^x}+\left(\frac{2010}{2011}\right)^x}{-\frac{x^{2010}}{2011^x} + 1}
$$
Where can you go from there?
Also, for the record: for any positive number $a$, we have
$$
\frac{d}{dx}a^x = \frac{d}{dx}e^{ln(a)\,x} = \ln(a)e^{\ln(a)\,x} = \ln(a) \,a^x
$$

In general: if you have to evaluate a limit of the form $\lim_{x\to\infty} \frac{x^n}{a^x}$ for positive number $a$ and integer $n$, you could note that
$$
\lim_{x\to\infty} \frac{x^n}{a^x} = \left(\lim_{x\to\infty} \frac{x}{a^{x/n}}\right)^n = \left(\lim_{x\to\infty} \frac{x}{e^{\frac{\ln(a)}{n}x}}\right)^n
$$
From there, you could use L'Hôpital's rule.
A: As said by Brian M. Scott, for large values of x, the exponentials dominate and the problem reduces to the limit of (2010 / 2011)^x when x goes to infinity. So, the result is zero (as confirmed by Wolfram Alpha). 
