Say a function $$f(t) = \frac{t^2\arctan t}{t-4}$$
Obviously, this has a vertical asymptote at $t = 4$. However, the oblique asymptote, if there is one, I can't seem to find.
What I do is I put the oblique asymptotes equation to $y = kt + m$. I can then say that $k = \lim_{t\rightarrow\pm\infty}\frac{f(t)}{t}$ however, this equals infinity as $\lim_{t\rightarrow\pm\infty}t\arctan t = +\infty$.
Can I now safely assume there is no oblique asymptote? That's what I assumed until Wolfram told me otherwise.
Answer, as explained below by Hippalectryon
The mistake I made in the above text was to divide every term in my function by $t$, which is obviously incorrect. The correct equation after dividing $f(t)$ with $t$ is:
$$\frac{f(t)}{t} = \frac{t^2\arctan t}{t(t-4)}$$
Therefore we get
$$\displaystyle k = \lim_{t\rightarrow\pm\infty}\frac{f(t)}{t}=\lim_{t\rightarrow\pm\infty}\frac{t^2\cdot\arctan(t)}{t\cdot(t-4)}=\lim_{t\rightarrow\infty}\arctan(t)=\pi/2$$
$\displaystyle k = \lim_{t\rightarrow\pm\infty}\frac{f(t)}{t}=\lim_{t\rightarrow\pm\infty}\frac{t^2\cdot\arctan(t)}{t\cdot(t-4)}=\lim_{t\rightarrow\infty}\arctan(t)=\pi/2$