Examples of Proofs Involving Limits: 3 Practical Examples (Plus More You’ll Actually Use)

Limit proofs look abstract, but they show up everywhere: in how we define derivatives, in how we know infinite series converge, and in the rigorous foundations of calculus that modern STEM degrees rely on.

Universities like MIT and Harvard still teach limits using the epsilon–delta definition in their analysis and honors calculus courses (see, for example, MIT’s open courseware notes on analysis at mit.edu). They don’t do this to torture students. They do it because once you understand a few examples of proofs involving limits, you can reuse the same moves in more advanced topics: continuity, differentiability, uniform convergence, and more.

Instead of starting with dry theory, we’ll start with three of the best examples you’re likely to see on homework or exams, then extend those ideas to several more real examples.

Example 1: Proving a linear limit using epsilon–delta

We’ll begin with a classic, friendly one. This is often the first example of an epsilon–delta limit proof students see.

Claim: Prove that
\(\displaystyle \lim_{x \to 2} (3x + 1) = 7.\)

Strategy

The informal idea is easy: if \(x\) is close to 2, then \(3x + 1\) is close to 7. The formal epsilon–delta definition says:

For every \(\varepsilon > 0\), there exists \(\delta > 0\) such that whenever \(0 < |x - 2| < \delta\), we have
[
| (3x + 1) - 7 | < \varepsilon.
]

We work backwards from the inequality we want, then forwards in the formal proof.

Backward work (scratch work)

[
|(3x + 1) - 7| = |3x - 6| = 3|x - 2|.
]

We want \(3|x - 2| < \varepsilon\). That happens if we make
[
|x - 2| < \frac{\varepsilon}{3}.
]

So a natural choice is \(\delta = \varepsilon/3\).

Formal proof

Let \(\varepsilon > 0\) be given. Choose \(\delta = \varepsilon/3\).

Now assume \(0 < |x - 2| < \delta\). Then
[
|(3x + 1) - 7| = 3|x - 2| < 3\delta = 3\left(\frac{\varepsilon}{3}\right) = \varepsilon.
]

This matches the epsilon–delta definition, so
\[\lim_{x \to 2} (3x + 1) = 7.\]

This is one of the cleanest examples of proofs involving limits: 3 practical examples we’ll build on, because the same pattern works for any linear function \(ax + b\).

Example 2: A slightly trickier limit: \(\lim_{x \to 1} x^2 = 1\)

Next, let’s take a quadratic. This is a standard example of a limit proof that forces you to control an extra factor.

Claim: Prove that
\(\displaystyle \lim_{x \to 1} x^2 = 1.\)

Backward work

We want: for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that if \(0 < |x - 1| < \delta\), then
[
|x^2 - 1| < \varepsilon.
]

Factor:
[
|x^2 - 1| = |x - 1||x + 1|.
]

The trouble is \(|x + 1|\) changes with \(x\). A classic trick: trap \(x\) in a small neighborhood so \(|x + 1|\) is bounded.

Let’s force \(|x - 1| < 1\). Then \(x\) lies between 0 and 2, so \(|x + 1| \le 3\). That gives
[
|x^2 - 1| = |x - 1||x + 1| \le 3|x - 1|.
]

To make this less than \(\varepsilon\), it’s enough to have
[
3|x - 1| < \varepsilon \quad \Rightarrow \quad |x - 1| < \frac{\varepsilon}{3}.
]

So we want both \(|x - 1| < 1\) and \(|x - 1| < \varepsilon/3\). We can guarantee both by choosing
[
\delta = \min\left{1, \frac{\varepsilon}{3}\right}.
]

Formal proof

Let \(\varepsilon > 0\) be given. Choose
[
\delta = \min\left{1, \frac{\varepsilon}{3}\right}.
]

Assume \(0 < |x - 1| < \delta\). Then \(|x - 1| < 1\), so \(0 < x < 2\). Hence
[
|x + 1| \le 3.
]

Now
[
|x^2 - 1| = |x - 1||x + 1| \le 3|x - 1| < 3\delta \le 3\left(\frac{\varepsilon}{3}\right) = \varepsilon.
]

Therefore, \(\displaystyle \lim_{x \to 1} x^2 = 1\).

This is one of the best examples to study carefully, because the “trap x so a factor is bounded” trick appears in many examples of proofs involving limits later in analysis.

Example 3: A sequence limit: \(\lim_{n\to\infty} \frac{1}{n} = 0\)

Now let’s switch from functions of a real variable to sequences. Textbooks and open courses (like those at harvard.edu) treat sequence limits with a nearly identical definition, just with \(n\) instead of \(x\).

Claim: Prove that
\(\displaystyle \lim_{n\to\infty} \frac{1}{n} = 0.\)

Strategy

For sequences, the definition is:

For every \(\varepsilon > 0\), there exists a natural number \(N\) such that whenever \(n \ge N\), we have
[
\left|\frac{1}{n} - 0\right| < \varepsilon.
]

Backward work

We want
[
\frac{1}{n} < \varepsilon \quad \Rightarrow \quad n > \frac{1}{\varepsilon}.
]

So a natural choice is to take any integer \(N\) larger than \(1/\varepsilon\). A standard clean pick:
[
N = \left\lceil \frac{1}{\varepsilon} \right\rceil.
]

Formal proof

Let \(\varepsilon > 0\) be given. Choose
[
N = \left\lceil \frac{1}{\varepsilon} \right\rceil.
]

If \(n \ge N\), then \(n \ge 1/\varepsilon\), so
[
0 < \frac{1}{n} \le \frac{1}{N} \le \varepsilon.
]

Thus
[
\left|\frac{1}{n} - 0\right| = \frac{1}{n} < \varepsilon.
]

Therefore, \(\displaystyle \lim_{n\to\infty} \frac{1}{n} = 0\).

This is the third in our examples of proofs involving limits: 3 practical examples, and it introduces the same “work backward, then formalize” pattern in a slightly different setting.

More real examples: how the same ideas repeat

To strengthen your skills, it helps to see several more real examples that use the same tricks in slightly different ways. These examples include absolute values, roots, and one-sided limits—exactly the kind of variety that shows up on modern exams and online problem sets in 2024–2025.

Example 4: Limit involving an absolute value

Claim: Prove that
\(\displaystyle \lim_{x \to 3} |x - 3| = 0.\)

This one is almost embarrassingly nice.

We want: for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that if \(0 < |x - 3| < \delta\), then
[
||x - 3| - 0| < \varepsilon.
]

But
[
||x - 3| - 0| = |x - 3|.
]

So the condition we want is simply
[
|x - 3| < \varepsilon.
]

That suggests choosing \(\delta = \varepsilon\).

Formal proof: Let \(\varepsilon > 0\). Take \(\delta = \varepsilon\). If \(0 < |x - 3| < \delta\), then
[
||x - 3| - 0| = |x - 3| < \delta = \varepsilon.
]

So \(\displaystyle \lim_{x \to 3} |x - 3| = 0\).

This is a nice warm-up example of a limit proof with absolute values that you can generalize to \(\lim_{x\to a} |x-a| = 0\).

Example 5: A square root limit

Claim: Prove that
\(\displaystyle \lim_{x \to 4} \sqrt{x} = 2.\)

This is a standard textbook favorite and one of the best examples to practice algebraic manipulation.

We need: for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that if \(0 < |x - 4| < \delta\), then
[
|\sqrt{x} - 2| < \varepsilon.
]

Backward work

A classic trick is to multiply and divide by the conjugate:
[
|\sqrt{x} - 2| = \frac{|\sqrt{x} - 2||\sqrt{x} + 2|}{|\sqrt{x} + 2|} = \frac{|x - 4|}{|\sqrt{x} + 2|}.
]

We want
[
\frac{|x - 4|}{|\sqrt{x} + 2|} < \varepsilon.
]

So we need a lower bound on \(|\sqrt{x} + 2|\). If we restrict \(x\) near 4, say by requiring \(|x - 4| < 1\), then \(3 < x < 5\), so \(\sqrt{x} > \sqrt{3}\). That gives
[
|\sqrt{x} + 2| > 2 + \sqrt{3} > 3.
]

Then
[
|\sqrt{x} - 2| = \frac{|x - 4|}{|\sqrt{x} + 2|} < \frac{|x - 4|}{3}.
]

To make this less than \(\varepsilon\), it’s enough to have
[
|x - 4| < 3\varepsilon.
]

We also needed \(|x - 4| < 1\), so we choose
[
\delta = \min{1, 3\varepsilon}.
]

Formal proof

Let \(\varepsilon > 0\) be given. Choose
[
\delta = \min{1, 3\varepsilon}.
]

Assume \(0 < |x - 4| < \delta\). Then \(|x - 4| < 1\), so \(3 < x < 5\) and \(\sqrt{x} > \sqrt{3}\). Hence
[
|\sqrt{x} + 2| > 2 + \sqrt{3} > 3.
]

Therefore
[
|\sqrt{x} - 2| = \frac{|x - 4|}{|\sqrt{x} + 2|} < \frac{|x - 4|}{3} < \frac{\delta}{3} \le \frac{3\varepsilon}{3} = \varepsilon.
]

So \(\displaystyle \lim_{x \to 4} \sqrt{x} = 2\).

Again, this fits right into our collection of examples of proofs involving limits that rely on bounding a denominator by restricting \(x\) to a small interval.

Example 6: One-sided limit at 0

One-sided limits are important in modern calculus courses, especially when defining derivatives from the right or left. They’re also emphasized in updated AP Calculus materials and online notes from universities like the University of Texas and others.

Claim: Prove that
\(\displaystyle \lim_{x \to 0^+} \sqrt{x} = 0.\)

The definition for a right-hand limit is: for every \(\varepsilon > 0\), there exists \(\delta > 0\) such that if \(0 < x - 0 < \delta\) (that is, \(0 < x < \delta\)), then
[
|\sqrt{x} - 0| < \varepsilon.
]

Backward work

We want \(\sqrt{x} < \varepsilon\). Squaring both sides (valid for nonnegative numbers):
[
0 < x < \varepsilon^2.
]

So we can choose \(\delta = \varepsilon^2\).

Formal proof

Let \(\varepsilon > 0\) be given. Take \(\delta = \varepsilon^2\).

Suppose \(0 < x < \delta\). Then \(0 < x < \varepsilon^2\), so taking square roots gives
[
0 < \sqrt{x} < \varepsilon.
]

Thus
[
|\sqrt{x} - 0| = \sqrt{x} < \varepsilon.
]

Therefore, \(\displaystyle \lim_{x \to 0^+} \sqrt{x} = 0\).

This is a clean example of a one-sided limit proof where a simple algebraic operation (squaring) does the heavy lifting.

Example 7: Another sequence limit: \(\lim_{n\to\infty} \frac{2n+1}{n} = 2\)

Let’s revisit sequences with a slightly more interesting expression.

Claim: Prove that
\(\displaystyle \lim_{n\to\infty} \frac{2n+1}{n} = 2.\)

Strategy

Rewrite the term:
[
\frac{2n+1}{n} = 2 + \frac{1}{n}.
]

We already know from an earlier example that \(\frac{1}{n} \to 0\). So intuitively, this sequence should approach 2.

Formally, we need: for every \(\varepsilon > 0\), there exists \(N\) such that if \(n \ge N\), then
[
\left|\frac{2n+1}{n} - 2\right| < \varepsilon.
]

But
[
\left|\frac{2n+1}{n} - 2\right| = \left|2 + \frac{1}{n} - 2\right| = \left|\frac{1}{n}\right| = \frac{1}{n}.
]

We’ve already solved that: choose \(N = \lceil 1/\varepsilon \rceil\) so that for all \(n \ge N\), \(1/n < \varepsilon\).

Formal proof: Let \(\varepsilon > 0\) be given. Set \(N = \lceil 1/\varepsilon \rceil\). For any \(n \ge N\),
[
\left|\frac{2n+1}{n} - 2\right| = \frac{1}{n} \le \frac{1}{N} \le \varepsilon.
]

So \(\displaystyle \lim_{n\to\infty} \frac{2n+1}{n} = 2\).

This is a nice bridge between earlier examples of proofs involving limits for sequences and algebraic manipulation of expressions.

How to approach new proofs involving limits on your own

By now we’ve walked through more than the promised examples of proofs involving limits: 3 practical examples. We’ve seen linear functions, quadratics, absolute values, roots, and sequences. Notice how the same patterns keep showing up:

• Work backward from \(|f(x) - L| < \varepsilon\) to find a condition on \(|x - a|\) or on \(n\).
• Trap the variable in a small interval (like \(|x - 1| < 1\)) to bound annoying factors.
• Use algebraic tricks: factoring, conjugates, rewriting expressions.
• Then write a clean forward proof using your chosen \(\delta\) or \(N\).

If you’d like more practice problems and explanations, many universities host free notes and exercises. For instance, you can find rigorous limit definitions and proofs in open resources from places like MIT OpenCourseWare and real analysis notes at various .edu sites.

FAQ: Common questions about examples of proofs involving limits

Q1. Where can I find more examples of proofs involving limits beyond these 3 practical examples?
Many university calculus and analysis courses publish free notes and problem sets online. MIT OpenCourseWare and similar resources at large public universities often include dozens of worked limit proofs. Searching for “epsilon delta limit examples site:.edu” will turn up high-quality PDFs and lecture notes.

Q2. What’s a good example of a limit proof that uses both left and right limits?
A classic exercise is to prove that \(\lim_{x\to 0} |x| = 0\) by separately proving the right-hand limit \(\lim_{x\to 0^+} |x| = 0\) and the left-hand limit \(\lim_{x\to 0^-} |x| = 0\). This extends the style of Example 4 and Example 6 and shows how one-sided limits work together.

Q3. How do these examples of proofs involving limits relate to derivatives?
The derivative is defined as a limit: \(f’(a) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}\), when this limit exists. All the epsilon–delta techniques you see in examples of limit proofs can be adapted to prove that specific functions are differentiable at certain points. Honors calculus and real analysis courses often include formal proofs of derivative rules using these ideas.

Q4. Are epsilon–delta proofs still relevant in 2024–2025 when we have powerful calculators and software?
Yes. While tools like graphing calculators and computer algebra systems can approximate limits numerically, the rigorous definitions are still central in university-level math, theoretical computer science, and parts of physics and engineering. Modern curricula and online platforms continue to emphasize conceptual understanding, not just computation.

Q5. What’s an accessible example of a sequence limit that’s slightly harder than 1/n?
A nice next step is to prove that \(\lim_{n\to\infty} \frac{n}{n+1} = 1\). You can rewrite \(\frac{n}{n+1} = 1 - \frac{1}{n+1}\), then use the same reasoning as in the earlier sequence examples. It’s a gentle way to practice turning a new sequence into one of the familiar examples of proofs involving limits you already understand.

If you can walk through each example above without getting stuck, you’re in good shape to handle most limit proofs in an introductory analysis or honors calculus course. The next step is to try writing a few on your own—start with functions like \(5x - 4\), \(x^2 + 2x\), or sequences like \((3 - 1/n)\)—and use these examples as a template until the process feels natural.