Compute the limit $$\lim_{x\to-\infty}{\frac{2x+7}{\sqrt{x^2+2x-1}}}$$
Here are my steps:
$$\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}=\lim_{x\to-\infty}\frac{2x+7}{\sqrt{x^2+2x-1}}\cdot\frac{-1/x}{-1/x}=\frac{\displaystyle \lim_{x\to-\infty}-2-\frac{7}{x}}{\displaystyle \lim_{x\to-\infty}\sqrt{\frac{x^2+2x-1}{x^2}}}=\frac{-2}{1}=-2$$
But I am not confident about $\displaystyle \lim_{x\to-\infty}{\sqrt\frac{x^2+2x-1}{x^2}}$. I just suppose that $x$ can be 'large enough' to ignore $+2x-1$.
On
To avoid confution the best way when we deal with limit $\to -\infty $ is to change variable and set $x=-y$ with $y\to +\infty$ then
$$\lim_{x\to-\infty}{\frac{2x+7}{\sqrt{x^2+2x-1}}}=\lim_{y\to+\infty}{\frac{-2y+7}{\sqrt{y^2-2y-1}}}=\lim_{y\to+\infty}{\frac{y}{y}\frac{-2+7/y}{\sqrt{1-2/y-1/y^2}}}=-2$$
You are right. Note that\begin{align}\lim_{x\to-\infty}\sqrt{\frac{x^2+2x-1}{x^2}}&=\sqrt{\lim_{x\to-\infty}\left(1+\frac2x-\frac1{x^2}\right)}\\&=\sqrt{1}\\&=1.\end{align}