An application of L'Hopital's rule, rational equation with exponential functions. I am always impressed by how helpful people can be on this forum when it comes to math questions.
I am currently misunderstanding how to solve the following limit:
$$ \lim_{x \to \infty} \frac{4^x - 5^x}{2^x + 4^x}. $$
I realize that the natural logarithm needs to be applied and that the numerator and denominator need to be derived before the limit can be taken. I am guessing that my confusion is rooted in an inadequate understanding of taking the derivatives of logarithms.
Any clarification would be greatly appreciated. 
Thanks! 
 A: Hint: $a^x = e^{\ln a^x} = e^{x \, \ln a}$, so $\frac{\mathrm{d}}{\mathrm{d}x} a^x $ is now quite straightforward to compute, isn't it? Furthermore, your limit becomes a linear combination of $e^x$ terms in both the numerator and denominator, so they cancel out.
A: L'Hopital's rule sometimes finds answers quickly and easily that could not otherwise be found without resorting to pretty drastic measures, but it gives little or no insight.
Compare $5^x$ with $4^x$.  Every time $x$ increases by $1$, the former gets multiplied by $5$ and the letter by $4$.  Do this a few times
$$
5\times5\times5\times5\times5 = 3125
$$
$$
4\times4\times4\times4\times 4 = 1024
$$
At this point the former is more than three times as big as the latter.  Take each of $5$ and $4$ as a factor ten times and the former product will be more than nine times as big as the latter.  Twenty times and it will be more than eighty times as big.  Make the exponent $x$ big enough and the other exponential expressions will be microscopic compared to $5^x$.
That should give you the answer to the whole problem.
