Proof about $f$ with slant asymptotes

Question

Let $f$ be a function defined on $\mathbb{R}$. We say $f$ is a candy cane at $\infty$ if there exists A, B $\in \mathbb{R}$ with A $\neq$ 0 such that $$ \lim_{x \to \infty} \frac{f(x)}{Ax + B} = 1$$ Is each of the following statements true or false? If true, prove it. If false, show it with a counterexample.
(a) IF $f$ has a slant asymptote at $\infty$ THEN $f$ is a candy cane at $\infty$.
(b) IF $f$ is a candy cane at $\infty$ THEN $f$ has a slant asymptote at $\infty$.

(a) This is what I have soo far:
If f has a slant asymptote at $\infty$, then $$ \lim_{x \to \infty} [f(x) - (mx + c)] = 0 $$ $$ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} mx + c $$ Substituting $\lim_{x \to \infty} mx+ c$ for $\lim_{x \to \infty} f(x)$ we get $$ \frac{\lim_{x \to \infty} mx + c}{\lim_{x \to \infty} Ax + B} = 1$$ Taking x as the dominant term we get $$\frac{\lim_{x \to \infty} m + \frac{c}{x}}{\lim_{x \to \infty} A + \frac{B}{x}} = 1$$ Since $\frac{c}{x}$ and $\frac{B}{x}$ go to zero since $c$ and $B$ are constants divided by a large number, we can conclude that $$\frac{m}{A} = 1$$ This is because the limit of a constant is a constant. Thus $$m = A$$ Then it must be true that f is a candy cane.

(b) For b, I'm kind of unsure
If f is a candy cane at $\infty$, then $$\frac{\lim_{x \to \infty} f(x)}{\lim_{x \to \infty} Ax + B} = 1$$ I want to say that since the limit value is 1, $f(x)$ must be a linear function as well. I can take three cases:
Case 1: when degree of $f(x)$ is $>$ 1.
Case 2: when degree of $f(x)$ is $<$ 1.
Case 3: when degree of $f(x)$ is equal to 1.
And I can then conclude that it must the case that $f(x) is a linear function and must have a slant asymptote.

Thank you for your time.

Answer 1

$$\lim_{x \to \infty} \frac{x+\ln (x)}{x} =1$$ This shows that $x+\ln(x)$ is a candy cane at $\infty$. $$\lim_{x \to \infty} {x+\ln (x)-mx-c}=\infty$$ if $m \leqslant 1$. $$\lim_{x \to \infty} {x+\ln (x)-mx-c}=-\infty$$ if $m>1$. Thus, $x+\ln(x)$ has no slant asymptote. This proves that claim (b) is false.

To show that (a) is true, if we have: $$\lim_{x \to \infty} f(x)-mx-c=0 \implies \lim_{x \to \infty} \frac{f(x)-mx-c}{mx+c}=0$$ $$\lim_{x \to \infty} \frac{f(x)}{mx+c}-1=0 \implies \lim_{x \to \infty} \frac{f(x)}{mx+c}=1$$ Now, take $A=m$ and $B=c$. Note that you couldn't directly transposed $mx+c$ to the other side and then divide since both limits are infinite in value. Thus, you must divide inside the limit itself first.

This shows that claim (a) is true.

Answer 2

For (a), you say that $$ \frac{\lim_{x \to \infty} mx + c}{\lim_{x \to \infty} Ax + B} \stackrel?= 1.\tag1$$

This is not OK. You have written a "fraction" in which the denominator goes to either $\infty$ or $-\infty$. Likewise (if $m\neq 0$) the numerator also goes to either $\infty$ or $-\infty$. This gives you an indeterminate form.

Your next formula is OK, $$\frac{\lim_{x \to \infty} m + \frac{c}{x}}{\lim_{x \to \infty} A + \frac{B}{x}} = 1,\tag2$$ because the two limits are both finite and the denominator is not zero. So the thing to do (in cases where you need to get to a formula such as $(2)$) is to go from $\lim_{x \to \infty}\frac{mx + c}{Ax + B} = 1$ to Equation $(2)$ without passing through Equation $(1).$ You can do this by dividing numerator and denominator by $x$ before you try to take the limit of each one separately. (Aside from the fact that Equation $(1)$ is not OK, dividing each of two limits by the limit variable is also questionable.)

However, in this case all of this is unnecessary. All you need to do is to "guess" that $A=m$ and $B=c$ and then show that $ \lim_{x \to \infty} [f(x) - (mx + c)] = 0 $ implies $$ \lim_{x \to \infty} \frac{f(x)}{mx + c} = 1 .$$

For (b), note that the only polynomials that are candy canes are the first-degree polynomials $ax + b$ where $a \neq 0.$ So for the question to be interesting at all, we have to deal with functions that are not polynomials, for example, $x + \sin x.$