Answer
$$\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})=1$$
Work Step by Step
$$A= \lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2-x})$$
We cannot examine the behavior of $(\sqrt{x^2+x}-\sqrt{x^2-x})$ as $x$ approaches $\infty$ right away, because it will lead to the unsolved situation of $\sqrt{\infty}-\sqrt{\infty}=\infty-\infty$, which we would try to avoid.
Instead, we would want to turn it into a rational function and apply the usual method, by doing the followings:
$$A=\lim_{x\to\infty}\Big[(\sqrt{x^2+x}-\sqrt{x^2-x})\times\frac{(\sqrt{x^2+x}+\sqrt{x^2-x})}{(\sqrt{x^2+x}+\sqrt{x^2-x})}\Big]$$
$$A=\lim_{x\to\infty}\frac{(x^2+x)-(x^2-x)}{(\sqrt{x^2+x}+\sqrt{x^2-x})}=\lim_{x\to\infty}\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}$$
Now we can divide both numerator and denominator by the highest degree of $x$ in the denominator, which is $x$:
$$A=\lim_{x\to\infty}\frac{2}{\frac{\sqrt{x^2+x}}{x}+\frac{\sqrt{x^2-x}}{x}}$$
However, we must be careful with the sign here when we try to put $x$ inside the square root:
- We know that $|x|=\sqrt{x^2}$. Here, since $x\to\infty$, we consider values of $x\gt0$, therefore, $x = \sqrt{x^2}$.
$$A=\lim_{x\to\infty}\frac{2}{\frac{\sqrt{x^2+x}}{\sqrt{x^2}}+\frac{\sqrt{x^2-x}}{\sqrt{x^2}}}=\lim_{x\to\infty}\frac{2}{\sqrt{1+\frac{1}{x}}+\sqrt{1-\frac{1}{x}}}$$
$$A=\frac{2}{\sqrt{1+0}+\sqrt{1-0}}=\frac{2}{1+1}=\frac{2}{2}=1$$