Proving Limits Using The Definition: Step-by-Step Guide

October 17, 2025 by TextBrain Team 56 views

Proving Limits Using the Definition: A Step-by-Step Guide

Hey guys! Today, we're diving into the exciting world of limits in calculus. Specifically, we're going to tackle how to prove limits using their formal definition. This might sound intimidating, but trust me, we'll break it down into manageable steps. We'll be looking at a few examples, so by the end of this, you'll be a pro at proving limits!

Understanding the Limit Definition

Before we jump into the problems, let's quickly recap the formal definition of a limit. This is the foundation for everything we'll be doing. The limit definition, often referred to as the epsilon-delta definition, states:

For a function f(x), the limit as x approaches c is L (written as lim x→c f(x) = L) if and only if for every ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε.

Okay, let's break down what this mouthful actually means:

ε (epsilon): This represents an arbitrarily small positive number. Think of it as how close we want f(x) to be to the limit L.
δ (delta): This is another small positive number. It represents how close x needs to be to c to ensure that f(x) is within ε of L.
|x - c| < δ: This means the distance between x and c is less than δ.
|f(x) - L| < ε: This means the distance between f(x) and L is less than ε.

In simpler terms, to prove a limit, we need to show that for any desired level of closeness (ε) to the limit L, we can find a range around c (defined by δ) such that all x values within that range will make f(x) close enough to L. It's like saying, "No matter how picky you are about how close you want to be to the limit, I can always find a neighborhood around the input value that gets you there."

The key to mastering limit proofs lies in understanding this definition thoroughly. So, make sure you're comfortable with the concepts of epsilon and delta before moving on. Got it? Awesome! Let's tackle our first example.

Example 1: lim (as n approaches infinity) (3n+1)/(5n+2) = 3/5

This is a classic example involving a limit at infinity. Our goal is to show that as n gets incredibly large, the expression (3n+1)/(5n+2) gets arbitrarily close to 3/5. To do this, we'll apply the definition of a limit at infinity. This is slightly different from the epsilon-delta definition for limits at a finite point, but the core idea remains the same. For limits at infinity, we need to show that for every ε > 0, there exists an N > 0 such that if n > N, then |(3n+1)/(5n+2) - 3/5| < ε.

Here’s how we can prove it, step-by-step:

Start with the inequality |(3n+1)/(5n+2) - 3/5| < ε: Our mission is to manipulate this inequality until we can isolate n and find a suitable value for N.
Simplify the expression: Find a common denominator and combine the fractions:

|(3n+1)/(5n+2) - 3/5| = |(5(3n+1) - 3(5n+2)) / (5(5n+2))|

= |(15n + 5 - 15n - 6) / (25n + 10)|

= |-1 / (25n + 10)|

= 1 / (25n + 10) (since the absolute value makes it positive)
Set up the inequality: Now we have 1 / (25n + 10) < ε.
Solve for n: Let's isolate n:

1 < ε(25n + 10)

1 < 25εn + 10ε

1 - 10ε < 25εn

n > (1 - 10ε) / (25ε)
Choose N: We need to find an N such that if n > N, the original inequality holds. From our work, we see that if we choose N = (1 - 10ε) / (25ε), it seems like we're on the right track. However, there's a slight catch! We need N to be positive. If ε is too large (specifically, if ε > 1/10), then (1 - 10ε) will be negative, making N negative. To avoid this, we need to ensure that our choice of N is always positive. A safe choice is to take the absolute value and add a constant to ensure positivity, or we can choose a simpler upper bound.

A more practical choice for N is N = max{0, (1 - 10ε) / (25ε)} if we want to be rigorous. But for simplicity in many cases, especially in exams, we often overlook this rigor and directly use N = (1 - 10ε) / (25ε) assuming ε is small enough.

However, a better and widely accepted approach is to find a simpler N that still works. Observe that 25n + 10 > 25n, so:

1 / (25n + 10) < 1 / (25n)

Therefore, if 1 / (25n) < ε, then 1 / (25n + 10) < ε will also be true. Solving 1 / (25n) < ε for n gives:

1 < 25εn

n > 1 / (25ε)

So, we can choose N = 1 / (25ε).
Write the formal proof: Now, let's put it all together in a formal proof:

Let ε > 0 be given. Choose N = 1 / (25ε). Then, for all n > N, we have:

|(3n+1)/(5n+2) - 3/5| = 1 / (25n + 10)

< 1 / (25n) (since 25n + 10 > 25n)

< 1 / (25 * (1 / (25ε))) (since n > N = 1 / (25ε))

= ε

Thus, by the definition of a limit, lim (as n approaches infinity) (3n+1)/(5n+2) = 3/5.

See? We did it! We successfully proved the limit using the definition. The key takeaway here is manipulating the inequality to find a suitable N in terms of ε.

Example 2: lim (as n approaches infinity) (n^2 +1)/(2n^2 +1) = 1/2

Alright, let's keep the momentum going! This time, we'll tackle another limit at infinity, but with slightly more complex expressions. The goal is the same: show that as n becomes incredibly large, (n^2 +1)/(2n^2 +1) gets closer and closer to 1/2. We will follow the same steps as before, just with different algebra.

Start with the inequality |(n^2 +1)/(2n^2 +1) - 1/2| < ε
Simplify the expression: Find a common denominator and combine the fractions:

|(n^2 +1)/(2n^2 +1) - 1/2| = |(2(n^2 +1) - (2n^2 +1)) / (2(2n^2 +1))|

= |(2n^2 + 2 - 2n^2 - 1) / (4n^2 + 2)|

= |1 / (4n^2 + 2)|

= 1 / (4n^2 + 2) (since it's already positive)
Set up the inequality: Now we have 1 / (4n^2 + 2) < ε.
Solve for n: Isolate n:

1 < ε(4n^2 + 2)

1 < 4εn^2 + 2ε

1 - 2ε < 4εn^2

n^2 > (1 - 2ε) / (4ε)

n > √((1 - 2ε) / (4ε))
Choose N: Again, we need to be careful about the case where (1 - 2ε) is negative. To avoid issues, we can use a similar approach as before, or, for simplicity, find a larger expression that is easier to work with. Notice that 4n^2 + 2 > 4n^2, so:

1 / (4n^2 + 2) < 1 / (4n^2)

Thus, if 1 / (4n^2) < ε, then 1 / (4n^2 + 2) < ε will also be true. Solving 1 / (4n^2) < ε for n:

1 < 4εn^2

n^2 > 1 / (4ε)

n > √(1 / (4ε)) = 1 / (2√ε)

So, we can choose N = 1 / (2√ε).
Write the formal proof: Let's assemble the proof:

Let ε > 0 be given. Choose N = 1 / (2√ε). Then, for all n > N, we have:

|(n^2 +1)/(2n^2 +1) - 1/2| = 1 / (4n^2 + 2)

< 1 / (4n^2) (since 4n^2 + 2 > 4n^2)

< 1 / (4 * (1 / (4ε))) (since n > N = 1 / (2√ε), then n^2 > 1 / (4ε))

= ε

Therefore, by the definition of a limit, lim (as n approaches infinity) (n^2 +1)/(2n^2 +1) = 1/2.

Fantastic! We've successfully navigated another limit proof. You're getting the hang of this, right?

Example 3: lim (as n approaches infinity) n^2 / 2^n = 0

This example introduces something new: an exponential term (2^n). This will require a slightly different approach in our simplification, but the core concept of squeezing the expression remains the same. We aim to show that as n grows without bound, the fraction n^2 / 2^n approaches zero. This is a very important limit to understand, as it demonstrates how exponential functions grow much faster than polynomial functions.

Start with the inequality |n^2 / 2^n - 0| < ε: This simplifies to |n^2 / 2^n| < ε, which is just n^2 / 2^n < ε (since it's always positive for n > 0).
The Challenge: Directly solving this inequality for n is tricky. We need a clever way to bound the exponential term. This is where inequalities come to our rescue! A very useful inequality for this type of problem is that for n greater than some value, 2^n grows faster than any polynomial. In particular, we can use the fact that for n ≥ 5, n^2 < 2^n. However, we need something more precise to solve our problem.

A stronger inequality (which is less obvious but extremely helpful here) is based on the binomial theorem. For n ≥ 4:

2^n = (1 + 1)^n = 1 + n + n(n-1)/2 + n(n-1)(n-2)/6 + ... + 1

Notice that the term n(n-1)(n-2)/6 is a cubic term in n. For sufficiently large n, this term will be much larger than n^2. Specifically, for n ≥ 4:

2^n > n(n-1)(n-2)/6

Now, we want to relate this to n^2. We want to find an N such that for n > N, n^2 / 2^n < ε. Let's manipulate the inequality:

n^2 / 2^n < n^2 / [n(n-1)(n-2)/6] (using the binomial inequality)

= 6n / [(n-1)(n-2)]

Now we want to make 6n / [(n-1)(n-2)] < ε. This is still not easy to solve directly, but we can simplify it further by bounding the denominator. For large n, (n-1) ≈ n and (n-2) ≈ n, so (n-1)(n-2) ≈ n^2. Thus:

6n / [(n-1)(n-2)] ≈ 6n / n^2 = 6/n

So, we want 6/n < ε, which means n > 6/ε.
Choose N: Based on the above reasoning, we can choose N = max{4, 6/ε}. We need the max to ensure that we satisfy both the binomial inequality condition (n ≥ 4) and the inequality we derived (n > 6/ε).
Write the formal proof: This is where we put it all together, making sure the logic flows correctly:

Let ε > 0 be given. Choose N = max{4, 6/ε}. Then, for all n > N, we have:

n^2 / 2^n < n^2 / [n(n-1)(n-2)/6] (since n > N ≥ 4, we can use the binomial inequality)

= 6n / [(n-1)(n-2)]

Now, for n > 3, (n-1) > n/2 and (n-2) > n/2. So:

6n / [(n-1)(n-2)] < 6n / (n/2 * n/2) = 6n / (n^2 / 4) = 24 / n

Since n > N ≥ 6/ε, we have 24/n < 24 / (6/ε) = 4ε. This is not exactly what we wanted (we wanted < ε), but it's close. We can adjust our choice of N slightly to fix this. Instead of N = max{4, 6/ε}, let's try N = max{4, 24/ε}. Then:

24 / n < 24 / (24/ε) = ε

So, with this new choice of N, we have:

n^2 / 2^n < ε

Thus, by the definition of a limit, lim (as n approaches infinity) n^2 / 2^n = 0.

Whoa! This one was a bit trickier, right? The key here was finding the right inequality to help us bound the exponential term. Don't worry if it didn't click right away; these types of proofs take practice.

Example 4: lim (as n approaches infinity) n^2/(n+1) = infinity

Now, let's switch gears a bit and tackle a limit that goes to infinity. This means we need a different definition. The definition for a limit going to infinity states that for every M > 0, there exists an N > 0 such that if n > N, then f(n) > M. In simpler terms, no matter how big a number (M) you give me, I can find a point (N) such that for all n beyond that point, the function values f(n) are even bigger than M.

Understand the Definition: We need to show that for any M > 0, we can find an N such that if n > N, then n^2/(n+1) > M.
Manipulate the Inequality: We want to isolate n in the inequality n^2/(n+1) > M.

n^2 > M(n+1)

n^2 > Mn + M

n^2 - Mn - M > 0
Solve the Quadratic Inequality: This is a quadratic inequality. We can find the roots of the quadratic equation n^2 - Mn - M = 0 using the quadratic formula:

n = [M ± √(M^2 + 4M)] / 2

Since we're looking for large values of n, we're interested in the positive root:

n > [M + √(M^2 + 4M)] / 2
Choose N: We can choose N = [M + √(M^2 + 4M)] / 2. However, this looks a bit complicated. We can find a simpler N by using the fact that √(M^2 + 4M) > M for M > 0. So:

[M + √(M^2 + 4M)] / 2 > [M + M] / 2 = M

Thus, N = M is a valid choice, though it's not the tightest bound. A simpler (though slightly less precise) choice is N = M. Notice that if n > 2M, then n/2 > M and therefore, n^2 > Mn + M will also be satisfied for large n values.
Write the Formal Proof: Let's put it all together:

Let M > 0 be given. Choose N = 2M. Then, for all n > N, we have:

n > 2M

n^2 > 2Mn

We want to show n^2 / (n+1) > M. Let's manipulate the inequality:

n^2 > M(n+1)

n^2 > Mn + M

Since n > 2M, n^2 > 2Mn, so it suffices to show that 2Mn > Mn + M:

2Mn > Mn + M

Mn > M

n > 1

Since we're considering large n, n > 1 is satisfied. Thus:

n^2 > Mn + M

n^2 / (n+1) > M

Therefore, by the definition of a limit going to infinity, lim (as n approaches infinity) n^2/(n+1) = infinity.

Woohoo! We successfully proved a limit going to infinity. The key here was understanding the definition and finding a suitable N based on M.

Key Takeaways and Practice Tips

Okay, guys, we've covered a lot of ground! Let's recap the key takeaways for proving limits using the definition:

Understand the Definition: Make sure you thoroughly grasp the epsilon-delta definition (for finite limits) and the definition for limits going to infinity.
Manipulate the Inequality: The heart of the proof lies in manipulating the inequality |f(x) - L| < ε (or f(n) > M for limits at infinity) to isolate x or n.
Choose the Right δ or N: This is where the magic happens! You need to find a δ (in terms of ε) or an N (in terms of M) that guarantees the inequality holds.
Write the Formal Proof: Clearly state your choice of δ or N and show step-by-step how the inequality leads to the conclusion.
Practice, Practice, Practice: The more you practice, the more comfortable you'll become with these proofs. Start with simpler examples and gradually work your way up to more complex ones.

Here are some tips for practicing:

Start with the Basics: Work through examples in your textbook or online resources.
Don't Be Afraid to Make Mistakes: Mistakes are learning opportunities! Analyze where you went wrong and try again.
Break Down the Problem: If you're stuck, try breaking the problem down into smaller steps.
Seek Help When Needed: Don't hesitate to ask your professor, TA, or classmates for help.

Conclusion

Proving limits using the definition can be challenging, but it's a fundamental skill in calculus. By understanding the definitions, practicing regularly, and breaking down problems into manageable steps, you can master this important concept. So, keep practicing, and you'll be proving limits like a pro in no time! You got this! Now go forth and conquer those limits!