Limits and One-Sided Behavior

The question a limit answers

Suppose a sensor reports $f(x)$ as its input approaches a target $a$ . A limit asks whether the outputs approach one number. The input need not reach $a$ , and the function need not even be defined there. This distinction is what makes limits useful for holes, instantaneous rates and infinite processes.

We write $\lim_{x\to a}f(x)=L$ when outputs can be made arbitrarily close to $L$ by taking inputs sufficiently close to, but different from, $a$ . A table is evidence, not a proof: a function can behave differently between the sampled inputs.

Approach from both sides

The left-hand limit uses $x<a$ ; the right-hand limit uses $x>a$ . A finite two-sided limit exists precisely when both one-sided limits exist and agree. The actual value $f(a)$ is a separate question.

For finite limits, sums and products follow their corresponding algebraic operations. A quotient limit is the quotient of the limits only when the denominator's limit is nonzero. Substitution into a continuous formula is efficient; substitution yielding $0/0$ asks for more reasoning, not division by zero.

Worked example: a removable hole

For $x\ne2$ , factor before taking the limit:

\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2.

Therefore the limit at $2$ is $4$ . Cancellation is valid for nearby inputs because they are not $2$ . It does not define the original expression at $2$ . Setting $f(2)=99$ would leave the limit unchanged; setting $f(2)=4$ would fill the hole continuously.

Worked example: a jump

Let $g(x)=-1$ when $x<0$ , $g(0)=0$ , and $g(x)=1$ when $x>0$ . Then the left limit is $-1$ and the right limit is $1$ . There is no two-sided limit, even though $g(0)$ exists. Averaging the one-sided values does not produce a limit.

Precision and common mistakes

In the formal definition, for every $\varepsilon>0$ there must be a $\delta>0$ such that $0<|x-a|<\delta$ implies $|f(x)-L|<\varepsilon$ . For $f(x)=3x+1$ at $a=2$ , choosing $\delta=\varepsilon/3$ works because $|f(x)-7|=3|x-2|$ . One choice must control all sufficiently nearby inputs, not just a chosen sequence.

Do not treat “undefined at the point” as “no limit.” Conversely, a plotted point does not establish a nearby trend. The graph's vertical scale can also hide a small jump.

Practice

Find $\lim_{x\to3}(x^2-9)/(x-3)$ .
Find the one-sided limits of $|x|/x$ at zero.
A function equals $x^2$ except that $f(1)=8$ . Find its limit at $1$ and decide whether it is continuous there.

Show worked solutions

For $x\ne3$ , cancel $x-3$ to obtain $x+3$ , so the limit is $6$ .
On the left the quotient is $-1$ ; on the right it is $1$ . The two-sided limit does not exist.
Nearby values follow $x^2$ , so the limit is $1$ . Continuity fails because $f(1)=8\ne1$ .

Explore

Approach the limit

Try this. Reduce the distance toward zero. The left and right outputs approach 4, even though the point at x = 2 is missing.

Distance from x = 20.5

f(x) = (x² − 4)/(x − 2), x ≠ 2. From the left: 3.5. From the right: 4.5. Both approach 4; f(2) is undefined.