<s>[INST] Calculating a limit. Is WolframAlpha wrong or am I wrong? What I'm trying to solve:
$$\lim _{x\to -\infty \:}\frac{\left(\sqrt{\left(x^2+14\right)}+x\right)}{\left(\sqrt{\left(x^2-2\right)}+x\right)}$$
What I put into WolframAlpha:
(sqrt(x^2+14)+x)/(sqrt(x^2-2)+x)
My result: $1$, which I get by simply dividing bot the numerator and the denominator by $x$, then letting it go towards $-\infty$
$$\frac{\frac{\left(\sqrt{x^2+14}+x\right)}{x}}{\frac{\left(\sqrt{x^2-2}+x\right)}{x}}=\frac{\left(\sqrt{\frac{x^2}{x^2}+\frac{14}{x^2}}+\frac{x}{x}\right)}{\left(\sqrt{\frac{x^2}{x^2}-\frac{2}{x^2}}+\frac{x}{x}\right)}=\frac{\left(\sqrt{1+\frac{14}{x^2}}+1\right)}{\left(\sqrt{1-\frac{2}{x^2}}+1\right)}\:=\:\frac{\left(\sqrt{1}+1\right)}{\left(\sqrt{1}+1\right)}\:=\:\frac22 \ = \ 1$$
WolframAlpha's result: $-7$. It has a long, complicated 25 step solution.
Is the solution $1$, $-7$ or neither?
Edit: of course, I set it $x$ to go towards $-\infty$ in WolframAlpha too
 [/INST] You can divide by $x$ but not in that way. Since $x\to-\infty$ it is eventually negative i.e. $-x>0$ so you should have $$\frac{\sqrt{x^2+c}}x = \frac{\sqrt{x^2+c}}{-\sqrt{x^2}}=  -\sqrt{1+\frac c{x^2}}=-1-\frac c{2x^2}+O(|x|^{-3})$$
This last inequality via the approximation $\sqrt{1+t}= 1 + \frac t2+O(t^2)$ which can be shown via Taylor’s theorem. Thus
$$\frac{\sqrt{x^2+14}+x} {\sqrt{x^2-2}+x}=\frac{-14x^{-2}/2 +O(|x|^{-3})}{2x^{-2}/2 + O(|x|^{-3})} \to -7 \qquad (x\to-\infty)$$
quick Desmos graph to verify:

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