0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.