Does the function $\frac{x}{\sqrt{1+x}}$ have an oblique asymptote? Does the function $$\frac{x}{\sqrt{1+x}}$$ have an oblique asymptote? If so, how do we find it? I thought about long dividing, but that wouldn't work, because You can't divide square roots. 
 A: A necessary condition for the function $f$ to have an oblique asymptote at $\infty$ is that
$$
\lim_{x\to\infty}\frac{f(x)}{x}
$$
exists, is finite and is non zero. If the oblique asymptote exists, then this limit is its slope. It mustn't be zero, because otherwise it wouldn't be oblique in the first place.

If an oblique asymptote exists, with equation $y=mx+q$ ($m\ne0$), then, by definition,
$$
\lim_{x\to\infty}(f(x)-mx-q)=0
$$
so also
$$
\lim_{x\to\infty}\frac{f(x)-mx-q}{x}=0
$$
and then
$$
\lim_{x\to\infty}\left(\frac{f(x)}{x}-m\right)=0
$$
because, of course, $\lim_{x\to\infty}\frac{q}{x}=0$.
A: HINT:
$$\frac{x}{\sqrt{1+x}}=\frac{x+1-1}{\sqrt{x+1}}=\sqrt{1+x}-\frac{1}{\sqrt{1+x}}$$
A: You are not going to have an "oblique" asymptote like you are used to--it's not going to be a straight line.  Instead the "oblique asymptote" is going to be a sideways parabola.  This is because you have (essentially) $\frac{x}{\sqrt{x}} = \sqrt{x}$--so the asymtptote is going to be a square root--or quadratic:
$$
f(x) = \frac{x}{\sqrt{x + 1}} = x\frac{\sqrt{x + 1}}{x + 1}
$$
Put another way what you have is: $\sqrt{x + 1} = y \rightarrow y^2 = x+1 \rightarrow x = y^2 - 1$:
$$
\left(y^2 - 1\left)\frac{y}{y^2}\right.\right. = \frac{y^2 - 1}{y} = y - \frac{1}{y} = \sqrt{x + 1} - \frac{1}{\sqrt{x + 1}}
$$
This means that the "oblique" asymptote is $y = \sqrt{x + 1}$.
