About a function $f$, of domain $\mathbb{R}$, it is known that it is odd and that $\lim_{x \rightarrow \infty} [f(x)+x-1]= 0$. Let $g$ be the restriction of $f$ to $\mathbb{R}^-$. Which of the following lines is the asymptote of the graph of function $g$?

A) $y = x-1$

B)$y = x+1$

C)$y = -x-1$

D)$y = -x+1$

I did:

$$\lim_{x \rightarrow \infty} [f(x)+x-1]= 0 \Leftrightarrow \lim_{x \rightarrow \infty} [f(x)-(-x+1)]= 0 \Leftrightarrow \lim_{x \rightarrow \infty} [f(x)]= -x +1$$

So one of the asymptotes of $f$ is $-x+1$. Now, because the function is odd, this excludes options A) and B).

But what do I do next? My book says the solution is C), but I don't understand why. Shouldn't it be D)?

Can someone explain this to me?


The last step in your derivation reads $\lim f(x) = - x + 1$; this is incorrect, and of course what you mean is that $f(x) \sim (1-x)$ for $x \uparrow +\infty$. (The symbol $\sim$ reads "is asymptotic to".)

Now, recall that $f(x)$ is odd if $f(x) = - f(-x)$ for all $x$. Then $f(x) \sim (1-x)$ for $x \uparrow +\infty$ turns into $-f(-x) \sim (1-x)$ for $x \uparrow +\infty$. Replacing $t=-x$, this becomes $-f(t) \sim (1+t)$ for $t \downarrow -\infty$ or, equivalently, $f(t) \sim -(1+t)$ for $t \downarrow -\infty$. So C is correct.


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