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

The limit is as $x$ goes to $-\infty,$ so you can't just divide by $x.$ Both the numerator and the denominator have the form $\infty -\infty.$ WolframAlpha is correct.
