I have to find the oblique asymptote for the following equation: $$y=\frac{2x^{2}-4x-1}{x-3}$$
I applied long division to bring it into the form of $Q + \frac{R}{D}$ leaving me with: $$y=2x+2+\frac{5}{x-3}$$
And to find the oblique asymptote I took the limit as $x$ approaches infinity of $y$, that left me with: $y=\infty$ however in the textbook I am studying it states that;
As $x$ increases then $\frac{5}{x-3}$ approaches $0$ so the equation of the oblique asymptote is $y=2x+2$
Why have we disregarded $2x+2$ in taking the limit, should it not also tend to infinity?
Best Answer
Let $f(x):= \frac{2x^{2}-4x-1}{x-3}$ and $g(x)=2x+2.$ Then:
$$|f(x)-g(x)|= \frac{5}{x-3}$$
for $x>3$.
Hence $f(x)-g(x) \to 0$ as $x \to \infty.$
Thus $f(x) \approx g(x)$ for large $x$.