Find the limit of the equation
$$\lim_{x\to-\infty} (x+\sqrt{x^2 + 2x})$$
I start by multiplying with the conjugate:
$$\lim_{x\to-\infty} \left[(x+\sqrt{x^2 + 2x})\left({x – \sqrt{x^2 + 2x}\over x – \sqrt{x^2+2x}}\right)\right]$$
$$\lim_{x\to-\infty} {x^2 – (x^2 + 2x)\over x – \sqrt{x^2+2x}}$$
$$\lim_{x\to-\infty} {-2x\over x – \sqrt{x^2+2x}}$$
divide by highest power of denominator
$$\lim_{x\to-\infty} {(\frac1x)(-2x)\over (\frac1x) x – ({1\over \sqrt{x^2}})\sqrt{x^2+2x}}$$
$$\lim_{x\to-\infty} {-2\over 1 – \sqrt{1+\frac2x}} = {-2 \over 1-\sqrt{1 + 0}} = {-2 \over 0}$$
but I know this is wrong as the answer is $-1$. Where did I mess up? Thanks.
Best Answer
Notice that since $x$ is tending to $-\infty$ then $x=-\sqrt{x^{2}}$. So when you let $x$ enter the square root in the denominator the minus that was originally there becomes a plus due to cancellation. That is:
$$\frac{-2}{\frac{x-\sqrt{x^{2}+2x}}{x}}=\frac{-2}{1-\frac{1}{x}\sqrt{x^{2}+2x}}=\frac{-2}{1+\sqrt{1+\frac{2}{x}}}$$
which tends to $-1$ as $x$ tends to $-\infty$.