calculus - Finding limits of rational functions as $xtoinfty$

$$\displaystyle \lim_{x\to\infty}\frac{(2x^2+1)^2}{(x-1)^2(x^2+x)}.$$

I do not know where to start on this, I tried multiplying it out and that didn't help really. It seems very complicated and I know I have to reduce it somehow but everything I do just makes it more complicated.

Answer

Hint: Divide the numerator and the denominator by the highest power of $x$ that occurs (in this case, $x^4$); distribute it so that it shows you exactly what is going on. For example,
$$\begin{align*} \frac{1}{x^4}(x-1)^2(x^2+x) &=\frac{1}{x^2}\times\frac{1}{x^2}\times(x-1)^2\times(x^2+x) &&\text{(factor }\frac{1}{x^4}\text{)}\\ &= \frac{1}{x^2}\times(x-1)^2 \times \frac{1}{x^2}\times (x^2+x) &&\text{(reordering)}\\ &= \left(\frac{1}{x^2}(x-1)^2\right)\times\left(\frac{1}{x^2}(x^2+x)\right)&&\text{(associativity)}\\ &= \left(\frac{1}{x}(x-1)\frac{1}{x}(x-1)\right)\times\left(\frac{1}{x^2}(x^2+x)\right)\\ &=\left(\frac{x-1}{x}\times\frac{x-1}{x}\right)\times\left(\frac{x^2+x}{x^2}\right)\\ &=\left(\frac{x-1}{x}\right)^2\left(\frac{x^2+x}{x^2}\right). \end{align*}$$
Now simplify a bit. Then do the same thing with the numerator, and see what happens as $x\to\infty$.

(Once you have more familiarity with end behavior and limits of rational functions, you'll be able to compute this limit "by eye").