Can't find limit tending to infinity of a sequence I'm stumped by
$$\lim_{x \to \infty}\frac{1+3+5+\cdots+(2x-1)}{x+3} - x$$
My obvious first step was to get a lowest common denominator by $x(\frac{x+3}{x+3})$, giving
$$\lim_{x \to \infty}\frac{1+3+5+\cdots+(2x-1)-x^2-3x}{x+3} $$
But from here I'm stumped, because with x tending to infinity, the $2x-1-x^2$ part of the numerator will be indeterminate, won't it?  I was hoping to calculate the answer via the highest powers on both sides of the fraction, which I know you can do when the variable tends to infinity, but then I'd get an answer of $-\infty$ which is incorrect according to my solution book.
What did I miss? 
In edit, thanks to those have responded so far, but I'm even more confused. Here's the solution in my answer book:
$$\lim_{x \to \infty}\frac{(1+2x-1)\frac{x}{2}}{x+3} - x $$
$$\lim_{x \to \infty}\frac{x^2-(x+3)x}{x+3} = -3 $$
Does this make sense to any of you?  You know your stuff, I'm willing to believe that either the question was badly worded or the answer is wrong.

Assuming $\;x\in\Bbb N\;$ (Otherwise I cannot understand the expression), we get:
$$\frac{1+3+5+\ldots+(2x-1)}{x+3}-x=\frac{x^2}{x+3}-x=-\frac{3x}{x+3}\xrightarrow[x\to\infty]{}-3$$
