# How can an oblique asymptote be $y = x$ , as $x\to \infty$?

In my Calculus book, an oblique asymptote defined as:

Oblique Asymptote:

the function $y = f(x)$ has an oblique asymptote $y = mx + n$, if:

1. $$\lim_{x\to \infty} {f(x) \over x} = m$$ where $m$ is a finite number.

2. $$\lim_{x\to \infty} [{f(x) - mx}] = n$$ where $n$ is a finite number as well.

My problem:

In the exercises book, there's an exercise:

$$f(x) = x + {\sin(x) \over x}$$

I found there's no vertical asymptote, as in the solutions.

But, I also found that there's no oblique asymptote, but the solutions tell the opposite.

My solution:

$$m = \lim_{x \to \infty} x + {\sin x \over x} = \lim_{x \to \infty} x + {0} = \lim_{x \to \infty} {x} = {\infty}.$$

As we can see, by his own definition, the asymptote couldn't be - because $m$ is infinity,

as the definition defines that $m$ and $n$ must be finite.

So how the oblique asymptote is $y = x$?

Thank you!

• It's $m = \lim_{x \rightarrow \infty} \frac {f(x)}x$, not $\lim_{x \rightarrow \infty} f(x)$. – Kaster Jun 27 '13 at 8:08
• @Kaster You're right, I'll fix it, but still - I can't understand how an horzintal asymptote can be infinity – Billie Jun 27 '13 at 8:09

We have $$m=\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}1+\frac{\sin x}{x^2}=1$$ and $$n=\lim_{x\to\infty}f(x)-x=\lim_{x\to\infty}\frac{\sin x}{x}=0$$ so $$y=x$$ is the oblique asymptote of the function at $\infty$
$m=\lim_{x\to\infty}\frac{f(x)}{x}=\lim_{x\to\infty}1+\frac{\sin x}{x^2}=1$. Now you can continue.