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
 A: 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.
A: 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:

A: If you let $t={1\over |x|}$ and divide top and bottom by $|x|,$
you get
$$\lim_{t\to 0^+}{\sqrt{1+14t^2}-1\over \sqrt{1-2t^2}-1},$$
to which you can apply l'Hôpital's rule.
A: Your statement is correct for the limit as $ \ x \ $ approaches "positive infinity",
$$\lim _{x \ \to \ +\infty \:}\frac{\left(\sqrt{\left(x^2+14\right)}+x\right)}{\left(\sqrt{\left(x^2-2\right)}+x\right)}  \ \ = \ \ 1 \ \ , $$
and the means by which you calculated it is valid in that direction.
A basic issue which arises in the expression for which we are taking limits is that the function $ \ \sqrt{x^2+a} \ $ (with any real $ \ a \ $) has even symmetry on its domain, while $ \ x \ $ has odd symmetry, so the numerator, denominator, and the rational function produced are not symmetric about the $ \ y-$ axis.  Rather than having a single horizontal asymptote, this function has two distinct asymptotes.
We can examine the general function $  \ x \ + \sqrt{x^2+a} \  . $  The asymptotes for $  \ \sqrt{x^2+a} \ $ are given by $ \ y \ = \ |x| \ , $ so as $ \ x  \rightarrow  +\infty \ , $ we expect that $  \ x \ + \ \sqrt{x^2+a}  \ \rightarrow \ 2x \ . $  So a ratio of two such functions has the limit
$$\lim _{x \ \to \ +\infty \:}\frac{\left( \ x \ + \ \sqrt{ x^2+a } \ \right)}{\left( \ x  \ + \ \sqrt{ x^2+b} \ \right)}  \ \ \rightarrow \ \ \frac{2x}{2x} \ \ =  \ \ 1 \ \ , $$
independently of the values of $ \ a \ $ and $ \ b \ , $ as you demonstrated.
For the limit as $ \ x  \rightarrow  -\infty \ , $ however, we can see that $  \ x \ + \ \sqrt{x^2+a}  \ \rightarrow \ x \ + \ |x| \ \rightarrow \ x \ - \ x \ \rightarrow \ 0 \ . $  But it does so in a way that does depend on the sign and value of $ \ a \ , $ since $ \sqrt{x^2+a}  \ $ approaches $ \ -x \ $ "from above" for $ \ a > 0 \ $ and "from below" for $ \ a < 0  \ \ . $  Applying the "conjugate-factor" method, we produce
$$ \lim _{x \ \to \ -\infty \:} \ \frac{( \ x \ + \ \sqrt{x^2+a} \ ) · ( \ x \ - \ \sqrt{x^2+a} \ ) }{ x \ - \ \sqrt{x^2+a} } \ \ = \ \ \lim _{x \ \to \ -\infty \:} \  \ \frac{ x^2 \ - \ (x^2+a)}{ x \ - \ \sqrt{x^2+a} } $$ $$ = \ \ \lim _{x \ \to \ -\infty \:} \  \ \frac{- \ a}{ x \ - \ \sqrt{x^2+a} } \ \ . $$
We can "extract" a factor of $ \ x \ $ in the denominator, but as mentioned more than once among the comments, because we have $ \ x < 0 \ , $ we must work "under the square-root" as $ \ x \ = \ - |x|  \ = \ - \sqrt{x^2} \ : $
$$   \lim _{x \ \to \ -\infty \:} \  \ \frac{- \ a}{ x \ - \ \sqrt{x^2+a} } \ \ = \ \ \lim _{x \ \to \ -\infty \:} \  \ \frac{- \ a}{ x · ( \ 1 \ - \ \left[ \frac{1}{-\sqrt{x^2}} \right] · \sqrt{x^2+a} \ ) } $$ $$ = \ \ \lim _{x \ \to \ -\infty \:} \  \ \frac{- \ a}{ x · ( \ 1 \ + \  \sqrt{\frac{x^2}{x^2}  +  \frac{a}{x^2}} \ ) } \ \ \rightarrow \ \ \frac{- \ a}{ x · ( \ 1 \ + \  \sqrt{1 \ + \ 0} \ ) } \  = \ -\frac{a}{2x} \ \ \rightarrow \ \ 0 \ \ . $$
The "negative-infinity" limit for $  \ x \ + \sqrt{x^2+a} \   $  leaves the value of $ \ a \ $ "exposed", so the corresponding limit for our ratio becomes
$$\lim _{x \ \to \ -\infty \:}\frac{\left( \ x \ + \ \sqrt{ x^2+a } \ \right)}{\left( \ x  \ + \ \sqrt{ x^2+b} \ \right)}  \ \ \rightarrow \ \ \frac{-a}{-b} \ · \ \frac{2x}{2x} \ \ = \ \ \frac{a}{b} \ \ . $$
For the particular function in question then,
$$\lim _{x \ \to \ -\infty \:}\frac{x \ + \  \sqrt{ x^2+14 } }{x \ + \ \sqrt{ x^2-2 } }  \ \ = \ \ \frac{14}{-2} \ \ =  \ -7 \ \ . $$
