Q.Examine whether $\displaystyle\lim_{x\to0}\dfrac{|x|}{x}$ exists, by finding the left-hand and right-hand limits.

Because $|x|$ is defined piecewise (equal to $x$ for $x\ge0$ and to $-x$ for $x<0$), the function behaves differently on the two sides of $0$, so we must check the one-sided limits separately.

Step 1 — Left-hand limit. For $x$ approaching $0$ from the left ($x<0$, so $x\to0^-$), by definition $|x|=-x$ (since $x$ is negative, $|x|$ is its positive counterpart). So $\dfrac{|x|}{x} = \dfrac{-x}{x} = -1$ for every such $x$, and thus $\lim_{x\to0^-}\dfrac{|x|}{x} = -1$.

Step 2 — Right-hand limit. For $x$ approaching $0$ from the right ($x>0$, so $x\to0^+$), $|x|=x$. So $\dfrac{|x|}{x}=\dfrac{x}{x}=1$ for every such $x$, and $\lim_{x\to0^+}\dfrac{|x|}{x}=1$.

Step 3 — Compare. Since $\text{LHL}=-1$ and $\text{RHL}=1$ are unequal, the two-sided limit $\lim_{x\to0}\dfrac{|x|}{x}$ does not exist. …

A common mistake is to simplify $\dfrac{|x|}{x}$ to $1$ everywhere by carelessly cancelling, ignoring that $|x|=-x$ (not $x$) when $x$ is negative. Another is to conclude the limit 'is $0$' by some kind of averaging of $-1$ and $1$ — a …