Do there exist, $\underset{x\to \infty }{\text{lim}}\frac{3 \left| x\right| +2 x}{7 \left| x\right| -5 x}$?

Question

Here is why I think the limit does not exists.

link to image if mathjax doesn't work

$\lim \limits_{x \to ∞ } \frac{3|x|+2x}{7|x|-5x}$ == $\lim \limits_{1/h \to 0 } \frac{3|1/h|+2(1/h)}{7|1/h|-5(1/h)}$

$\lim \limits_{1/h \to 0^+ } \frac{3|1/h|+2(1/h)}{7|1/h|-5(1/h)}$ == $\lim \limits_{1/h \to 0^+ } \frac{3+2}{7-5}$ == $\frac{5}{2}$

$\lim \limits_{1/h \to 0^- } \frac{3|1/h|+2(1/h)}{7|1/h|-5(1/h)}$ == $\lim \limits_{1/h \to 0^- } \frac{-3+2}{-7-5}$ == $\frac{1}{12}$

left limit does not match the right limit. so i think the limit does not exist. Is this solution correct?

Edit: please note here the question is whether the limit to the given equation exists or not, if it does what is it.

Answer 1

$$\underset{x\to +\infty }{\text{lim}}\frac{3 \left| x\right| +2 x}{7 \left| x\right| -5 x}=\underset{x\to +\infty }{\text{lim}}\frac{3 x +2 x}{7x -5 x}=\underset{x\to +\infty }{\text{lim}}\frac{5x}{2x}=\frac{5}{2}$$ $$\underset{x\to -\infty }{\text{lim}}\frac{3 \left| x\right| +2 x}{7 \left| x\right| -5 x}=\underset{x\to -\infty }{\text{lim}}\frac{-3 x +2 x}{-7x -5 x}=\underset{x\to -\infty }{\text{lim}}\frac{-x}{-12x}=\frac{1}{12}$$ The limits at $\infty$ are two and they both exist. The limit at $x=0$ does not exist.

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Answer 2

When I started studying calculus, there still were books that distinguished between $+\infty$, $-\infty$ and $\infty$ (unsigned infinity).

The definition of $\smash{\lim\limits_{x\to\infty}}\,f(x)=l$ at unsigned infinity was

for every $\varepsilon>0$, there exists $M>0$ such that, for all $|x|>M$, it holds that $|f(x)-l|<\varepsilon$.

I don't think this has ever been useful, and nowadays it's quite rare to find it. Usually $\infty$ means what in those times was denoted as $+\infty$. Why is it useless? Because that (useless) limit exists if and only if the limits at $+\infty$ and $-\infty$ exist and they're equal. Which happens for rational functions when the degree of the numerator doesn't exceed the degree of the denominator: not so a general context.

In your case, you have $$ f(x)=\begin{cases} 5/2 & x>0 \\[6px] 1/12 & x<0 \end{cases} $$ so there's no doubt what the limit at positive and negative infinity are. Unless your textbook or instructor uses the old-fashioned unsigned infinity.

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The “unsigned infinity” has its uses, though: when doing with analytic functions, it is common to “compactify” the complex plane by adding a single point, namely $\infty$. And the notion of limit at $\infty$ is exactly the same as the one spelled above, but with the variable in the complex numbers.