Find all asymptotes of y=(x+1)/(sqrt(x²+2))

Find all asymptotes of $y=\dfrac{x+1}{\sqrt{x^2+2}}$.

Step 1: No vertical asymptote

$\sqrt{x^2+2}\ge\sqrt2>0$ for every real $x$, so the function is defined and finite everywhere.

Step 2: As $x\to+\infty$

Factoring $x$ out (with $\sqrt{x^2}=x$ for $x>0$): \[\lim_{x\to+\infty}\frac{x+1}{\sqrt{x^2+2}} = \lim_{x\to+\infty}\frac{1+1/x}{\sqrt{1+2/x^2}} = 1. \tag{1}\]

Step 3: As $x\to-\infty$

Here $\sqrt{x^2}=-x$, so

\[ \lim_{x\to-\infty}\frac{x+1}{\sqrt{x^2+2}} = \lim_{x\to-\infty}\frac{1+1/x}{-\sqrt{1+2/x^2}} = -1. \tag{2} \]

Step 4: Verification

\[x=200\text{: }y\approx0.99749;\qquad x=-200\text{: }y\approx-0.99749\ \checkmark\]