I have been told that
$f(x)$ has a slant right asynthote $m_1x+q_1$
$g(x)$ has a slant right asynthote $m_2x+q_2$
$m_1,m_2 \neq 0$
They ask me if $f(x) \sim g(x)$ when $x$ goes to infinity.
I don't even know what the question means. I made the following reasoning BUT I feel that something is not right with it:
1 means that
$$\lim_{x\to+\infty}(f(x) - m_1x+q_1)= 0$$
that can be written as
$$\lim_{x\to+\infty} f(x) = \lim_{x\to+\infty}( m_1x+q_1)$$
the same goes for $g(x)$ hence: $$\lim_{x\to+\infty} g(x) = \lim_{x\to+\infty}( m_2x+q_2)$$
I claim that $$\dfrac{\lim_{x\to+\infty} f(x)}{\lim_{x\to+\infty} g(x)}=\lim_{x\to+\infty} \dfrac{f(x)}{g(x)} = \dfrac{\lim_{x\to+\infty} m_1x+q_1}{\lim_{x\to+\infty} m_2x+q_2}=\lim_{x\to+\infty} \dfrac{m_1x+q_1}{m_2x+q_2} = \dfrac{m1}{m2}$$
I have the feeling that this is wrong. Could you please help me find out where the reasoning is wrong?
Thank you
EDIT:
Would it be possible to show that $f_(x) + 2g(x)$ has horizontal asymptote instead? I don't really know how to procede.
You're pretty close (and get the right answer), but your approach not quite correct, since $\lim_{x\to+\infty}f(x)=\lim_{x\to+\infty}(m_1x+q_1)$ is infinite, and the limit manipulation rules don't all work with infinite limits. Thus, for example, it's incorrect to state (as you do) that $\lim_{x\to+\infty}\frac{f(x)}{g(x)} = \frac{\lim_{x\to+\infty}f(x)}{\lim_{x\to+\infty}g(x)}$; the right side evaluates as $\frac{\infty}\infty$ (one or both infinities might be positive or negative), which has no defined value, while the left side might have a defined value (e.g., $1$, if $f=g$).
Instead, starting from $\lim_{x\to\infty}(f(x)-(m_1x+q_1))=0$, we have $\lim_{x\to\infty}\frac{f(x)-(m_1x+q_1)}{m_1x+q_1} = \lim_{x\to\infty} \left(\frac{f(x)}{m_1x+q_1}-1\right) = 0$, so $\lim_{x\to\infty} \frac{f(x)}{m_1x+q_1} = 1$. Similarly $\lim_{x\to\infty} \frac{g(x)}{m_2x+q_2} = 1$. Now take the reciprocal of the latter limit and multiply it by the first. See if you can continue from there.