Q.A function $f(x)$ is defined as $f(x) = x+1$ for $x<2$, and $f(x) = x^2$ for $x \geq 2$. Find the left-hand limit and the right-hand limit of $f(x)$ as $x \to 2$, and state whether $\lim_{x \to 2} f(x)$ exists.

Because $f$ is defined by two different formulas on either side of $x=2$, its tendency must be checked separately from the left and from the right before deciding whether an overall limit exists.

Approaching from the left uses the formula $x+1$, giving $2+1=3$; approaching from the right uses the formula $x^2$, giving $2^2=4$. Since $3 \neq 4$, the two-sided limit fails to exist at $x=2$.

For $x$ approaching $2$ from the left (values slightly less than $2$, e.g. $1.9, 1.99, 1.999,\ldots$), $x<2$ applies, so we use $f(x)=x+1$: $\lim_{x \to 2^{-}} f(x) = \lim_{x \to 2^{-}} (x+1) = 2+1 = 3$

For $x$ approaching $2$ from the right (values slightly greater than $2$, e.g. $2.1, 2.01, 2.001,\ldots$), $x \geq 2$ applies, so we use $f(x) = x^2$: $\lim_{x \to 2^{+}} f(x) = \lim_{x \to 2^{+}} x^2 = 2^2 = 4$

The left-hand limit is $3$ and the right-hand limit is $4$. Since a two-sided limit exists only when LHL $=$ RHL, and here $3 \neq 4$, we conclude: $\lim_{x \to 2} f(x) \text{ does not exist.}$

Note that this has nothing to do with whether $f(2)$ itself is defined — here $f(2) = 2^2 = 4$ is perfectly well defined (since the second formula applies at $x=2$ itself), yet the two-sided limit still fails to exist because the function approaches different values from the two sides.

A common error is substituting $x=2$ into both formulas and averaging, or simply picking whichever formula 'looks more natural' — the correct rule is to use the formula that actually applies on each side (values $<2$ use the first piece, values $\geq 2$ use the second), and both one-sided limits must be computed separately before comparing them.