<s>[INST] Which answer is correct? Finding the limit of a radical as $x$ approaches infinity. When I take $$\lim_{x \to -∞} \sqrt{x^2+7x}+x,$$ I multiply by the conjugate over the conjugate to get
$$\lim_{x \to -∞}\frac{7x}{\sqrt{x^2+7x}-x},$$ and multiply by either $\frac{\frac{1}{x}}{\frac{1}{x}}$ or $\frac{\frac{1}{-x}}{\frac{1}{-x}}$ to get an undefined answer or $\frac{-7}{2}.$
My teacher's solution involves multiplying by $\frac{\frac{1}{-x}}{\frac{1}{-x}}:$
$$=\lim_{x \to -∞}\frac{-7}{\sqrt{x^2/x^2+7x/x^2}+1}$$
$$=-\frac{7}{\sqrt{1+0}+1}$$
$$=\frac{-7}{2}$$
However, I multiplied
by $\frac{\frac{1}{x}}{\frac{1}{x}}$ and got the following:
$$\lim_{x \to -∞}\frac{7}{\sqrt{x^2/x^2+7x/x^2}-1}$$
$$\frac{7}{\sqrt{1+0}-1}$$
$$\frac{7}{0}$$
Which is undefined.
Why does multiplying by what is essentially $1$ cause different answers in general, and in particular for evaluating limits?
 [/INST] $$\lim_{x \to -\infty} \sqrt{x^2+7x}+x=$$
$$=\lim_{x \to -\infty} |x|\sqrt{1+\frac{7}{x}}+x=$$
$$=\lim_{x \to -\infty} -x\sqrt{1+\frac{7}{x}}+x=$$
$$=\lim_{x \to -\infty} -x\Big(\sqrt{1+\frac{7}{x}}-1\Big)=$$
$$=\lim_{x \to -\infty} -x\Big(1+\frac{7}{2x}+O\Big(\frac{1}{x^2}\Big)-1\Big)=$$
$$=\lim_{x \to -\infty} -\frac{7}{2}+O\Big(\frac{1}{x}\Big)=-\frac{7}{2}$$
</s>