Apparently in this form the limit is "trivially" 1/2 I'll admit, I'm struggling to keep up with the material in my analysis classes, but it only demoralizes me more when the questions in my textbooks have explained solutions, but I don't even understand how you go from one step to another.
This is the final from, that's supposed to be close to the simplest form (so close that it's "trivial" for the student to find it):
$\lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+\left(n-1\right)^{c+1}}{\left(c+1\right)\left(n^c-\left(n-1\right)^c\right)}\right)=\frac{1}{2}$
Why is this true? I don't see it at all.
 A: The $n^{c+1}$ and $n^c$ powers in both halves of the fraction are zero by design. The highest non-zero power is $n^{c-1}$:
$$\frac{\binom{c+1}2n^{c-1}+\cdots}{(c+1)\binom c1n^{c-1}+\cdots}$$
As $n\to\infty$ these terms will dominate, so the limit is
$$\frac{\binom{c+1}2}{(c+1)\binom c1}=\frac12$$
A: Compute the leading terms of binomial expansions in numerator and denominator and simplify
$$
\begin{align}
& \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+\left(n-1\right)^{c+1}}{\left(c+1\right)\left(n^c-\left(n-1\right)^c\right)}\right)= \\
& = \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+n^{c+1}-\binom{c+1}1 n^c + \binom{c+1}2 n^{c-1} + O(n^{c-2})}{\left(c+1\right)\left(n^c-n^c + \binom{c}1 n^{c-1}+O(n^{c-2})\right)}\right)\\
& = \lim _{n\to \infty }\left(\frac{\left(c+1\right)n^c-n^{c+1}+n^{c+1}-(c+1) n^c + \frac{c(c+1)}{2} n^{c-1} + O(n^{c-2})}{\left(c+1\right)\left(n^c-n^c + c n^{c-1}+O(n^{c-2})\right)}\right)\\
& = \lim _{n\to \infty }\left(\frac{\frac{c(c+1)}{2} n^{c-1} + O(n^{c-2})}{c\left(c+1\right)n^{c-1}+O(n^{c-2})}\right)\\
& = \frac{c(c+1)}{2}\frac{1}{c(c+1)} \\
& = \frac{1}{2}.
\end{align}
$$
