<s>[INST] Find limit of $\frac{2x+7}{\sqrt{x^2+2x-1}}$ (check my steps please..) 
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 $+2x-1$.
 [/INST] 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
$$\lim_{x\to-\infty}{\frac{2x+7}{\sqrt{x^2+2x-1}}}=\lim_{y\to+\infty}{\frac{-2y+7}{\sqrt{y^2-2y-1}}}=\lim_{y\to+\infty}{\frac{y}{y}\frac{-2+7/y}{\sqrt{1-2/y-1/y^2}}}=-2$$
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