Q.Find $\lim_{x\to 0} f(x)$, where $f(x) = \begin{cases} \dfrac{x}{|x|}, & x \neq 0 \\ 0, & x = 0 \end{cases}$

The key idea is that the limit depends on the sign of $x$ because of the absolute value in the denominator. For $x \neq 0$, $|x| = x$ when $x > 0$ and $|x| = -x$ when $x < 0$.

• For $x \to 0^+$ (right-hand limit):

  $\frac{x}{|x|} = \frac{x}{x} = 1$, so $\lim_{x\to 0^+} f(x) = 1$.

• For $x \to 0^-$ (left-hand limit):

  $\frac{x}{|x|} = \frac{x}{-x} = -1$, so $\lim_{x\to 0^-} f(x) = -1$. …

Why this problem matters

This is a classic example that separates the value of a function at a point from its limit at that point. Many students see $f(0)=0$ and assume the limit is also $0$ — but the limit depends entirely on the behaviour of $f(x)$ near $x=0$, not at $x=0$.

The function here is the sign function (or signum) in disguise. For $x \neq 0$, $x/|x|$ simply tells you the sign of $x$: $+1$ if $x>0$, $-1$ if $x<0$. At $x=0$, the function is artificially set to $0$.

Step-by-step reasoning

1. Understand the piecewise definition

   For $x \neq 0$, $f(x) = \dfrac{x}{|x|}$.

   • If $x > 0$, then $|x| = x$, so $f(x) = \dfrac{x}{x} = 1$.
   • If $x < 0$, then $|x| = -x$, so $f(x) = \dfrac{x}{-x} = -1$. At $x = 0$, the function is defined separately as $f(0) = 0$.

2. Recall the definition of a two-sided limit

   $\displaystyle \lim_{x\to 0} f(x)$ exists if and only if both one-sided limits exist and are equal:

3. Compute the right-hand limit ($x \to 0^+$) When $x$ approaches $0$ from the positive side, $x > 0$, so $f(x) = 1$ for every such $x$. Hence:

4. Compute the left-hand limit ($x \to 0^-$) When $x$ approaches $0$ from the negative side, $x < 0$, so $f(x) = -1$ for every such $x$. Hence:

5. Compare the two one-sided limits …