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I use to simplify the functions to eliminate factors when I want to find their limit at a specific value. Like in this post

But what do you do when the limit is something like this:

$$\lim_{x \to 0} \frac{|x|}{x}?$$

IS there a way to formally evaluate this? Or do you just rely on intuition? This function can't really be algebraically turned into another function like the one in my other post and so you can't get rid of its indeterminate form. It seems like you just have to evaluate it by looking at the two one sided limits and figuring out what $|x|$ is equal to when $x < 0$ or when $x > 0.$

When $x < 0,$ $$\frac{|h|}{h} = \frac{-h}{h} = - 1$$ while when $x > 0,$ it is $1.$ So the limit does not exist. But is there any way other than this to evaluate formally this limit?

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    $\begingroup$ Isn't it the formal way? Btw I don't know of any other method. $\endgroup$
    – jonsno
    Jan 8, 2018 at 7:18
  • $\begingroup$ As one of the answers says, use definition. To show it doesn't exist, you need to prove negation of it, normally $\forall x \exists y $ but you need to show that for all y there exists x such that... See negation of quantifiers. $\endgroup$ Jan 8, 2018 at 10:50

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Use a definition. It says a limit at $x_0$ is such value, to which function values $f(x)$ converge when the argument $x$ becomes close enough to $x_0$.

Formally: IF there exists such $L$ that for each (arbitrarily small) $\epsilon$ there exists such (small enough) $\delta$, that for each $x$ not farther from $x_0$ than $\delta$, values $f(x)$ are not farther than $\epsilon$ from $L$.

For your function $x\mapsto \frac{|x|}x$, as you pointed out, for each negative $x$ (that means also negative $x$ arbitrarily close to $0$) the function value is $-1$ while for each positive $x$ it is $1$. So for any $\delta$ there exist $x$-es in $(0-\delta, 0+\delta)$ which make $f(x)$ equal $-1$ and those which make $f(x)=1$. That implies e.g. for $\epsilon = 0.1$ there is no such $\delta$–neighborhood of $0$ which would make $f(x)$ fit into $(L-\epsilon, L+\epsilon)$ for any $L$, hence a requested $L$ $$L=\lim_{x\to0}\frac{|x|}x$$ does not exist.

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What “algebraic simplification” would you do for $$ \lim_{x\to 0}\frac{\sin x}{x} $$ or other similar limits?

Limits that can be computed with an algebraic simplification like the post you cite are basically derivatives of polynomials or of inverse functions thereof (via rationalization, maybe quite complicated).

Derivatives of polynomials can be defined algebraically and no concept of limit is needed for them. But functions are much more than polynomials.

And, no, there is no “general method” for deciding upon limits.

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The sign function can be defined as

$$\text{sgn}(x)=\frac{|x|}x=\frac x{|x|}$$ for $x\ne0$.

The function is piecewise constant, $-1$ on the left, $1$ on the right. To evaluate limits, the value at $0$ doesn't matter (you can set $\text{sgn(0)}:=0$).

Then you can use the rules

$$x=|x|\text{ sgn}(x),$$ $$|x|=x\text{ sgn}(x),$$ $$|\text{ sgn}(x)|=1^*.$$

$^*$Except at $x=0$.


Application:

$$\lim_{x\to0}\frac{|x|\sin|x|}{x^2}=\lim_{x\to0}\frac{\text{sgn}(x)|\sin x|}{x}=\lim_{x\to0}\frac{\text{sgn}(x)}{\text{sgn}(x)}\left|\frac{\sin x}{x}\right|=1\cdot1.$$

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You can use the definition of limit to show that the limit in question does not exist. Or you can find two sequences $a_n, b_n$ both tending to $0$ and $f(a_n) \to 1$ and $f(b_n) \to - 1$ so that the limit doesn't exist. Just choose $a_n=1/n,b_n=-1/n$.

If you want one standard procedure to evaluate all limits (like quadratic formula to solve a quadratic equation) then that's not possible. More importantly one has to get rid of desire to have one standard procedure for all problems.

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