Proving Limits Using Epsilon-Delta: A Complete Guide

Proving Limits with the Epsilon-Delta Definition: A Comprehensive Guide

Hey guys! Let's dive into the nitty-gritty of proving limits using the $\epsilon-\delta$ definition. This is a fundamental concept in calculus, and while it might seem a bit intimidating at first, we'll break it down step-by-step to make it super clear. Our goal? To prove that the limit of $\frac{3}{\sqrt{x+2}+4}$ as $x$ approaches 2 is equal to $\frac{1}{2}$. Buckle up; it's going to be a fun ride!

Firstly, what exactly is the $\epsilon-\delta$ definition? Well, it's the rigorous way of defining what a limit means. Forget informal descriptions; this is the backbone of limit proofs. The definition states that for any \epsilon > 0, there must exist a \delta > 0 such that if 0 < |x - a| < \delta, then |f(x) - L| < \epsilon. Here's the breakdown:

• $\epsilon$ (epsilon) represents a small positive number, essentially how close we want $f(x)$ to be to the limit $L$. Think of it as the margin of error.
• $\delta$ (delta) is also a small positive number, and it tells us how close $x$ needs to be to $a$ (the value $x$ approaches) to ensure that $f(x)$ is within the $\epsilon$ margin of error of $L$.
• $a$ is the value that $x$ is approaching (in our case, 2).
• $L$ is the limit we are trying to prove (in our case, $\frac{1}{2}$).
• $f(x)$ is the function whose limit we are trying to find (in our case, $\frac{3}{\sqrt{x+2}+4}$).

Basically, the definition says that no matter how small you make the error margin ($\epsilon$), you can always find a distance ($\delta$) around the point $a$ such that the function's values stay within that error margin of the limit $L$. This ensures that as $x$ gets closer to $a$, $f(x)$ gets closer to $L$.

Now, let's apply this to our specific problem. We want to show that:

$$\lim_{x\to 2}\frac{3}{\sqrt{x+2}+4}=\frac{1}{2}$$

So, we need to show that for any \epsilon > 0, there exists a \delta > 0 such that if 0 < |x - 2| < \delta, then |\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}| < \epsilon. Let's get started. To begin with, our aim is to find a relationship between $|x - 2|$ and $|\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}|$. The end goal of the proof is to make the expression $|\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}|$ less than $\epsilon$ by carefully selecting a $\delta$. This is done by manipulating the expression $|\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}|$ and relating it to $|x - 2|$.

Finding the Relationship: A Step-by-Step Approach

Alright, let's start by simplifying the expression $|\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}|$. This is where the magic happens. We will try to manipulate this expression to somehow get an $|x - 2|$ in there.

1. Combine Fractions: First, let's combine the fractions:

   $$|\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}| = |\frac{6 - (\sqrt{x+2}+4)}{2(\sqrt{x+2}+4)}| = |\frac{2 - \sqrt{x+2}}{2(\sqrt{x+2}+4)}|$$

2. Rationalize the Numerator: To get rid of the square root, let's rationalize the numerator. Multiply the numerator and denominator by the conjugate of the numerator, which is $2 + \sqrt{x+2}$.

   $$|\frac{2 - \sqrt{x+2}}{2(\sqrt{x+2}+4)}| = |\frac{(2 - \sqrt{x+2})(2 + \sqrt{x+2})}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})}| = |\frac{4 - (x+2)}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})}| = |\frac{2 - x}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})}|$$

3. Simplify and Isolate $|x - 2|$: Now, we can rewrite this as:

   $$|\frac{2 - x}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})}| = |\frac{-(x - 2)}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})}| = \frac{|x - 2|}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})}$$

   We've successfully isolated $|x - 2|$! The next step is to find a suitable $\delta$.

Notice how we have the $|x-2|$ term, which is what we want, but we also have other terms. The goal is to get rid of those additional terms or put a bound on them.

Bounding the Expression and Finding Delta

Now we want to choose a $\delta$ such that if 0 < |x - 2| < \delta, then \frac{|x - 2|}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})} < \epsilon. To do this, we need to find a bound for the denominator. Let's assume, for now, that $\delta \le 1$. This means that |x - 2| < 1, which implies that -1 < x - 2 < 1, or 1 < x < 3.

• Bounding $\sqrt{x+2}$: Since 1 < x < 3, we have 3 < x + 2 < 5. Therefore, \sqrt{3} < \sqrt{x+2} < \sqrt{5}.
• Bounding $2 + \sqrt{x+2}$: Since \sqrt{3} < \sqrt{x+2} < \sqrt{5}, we get 2 + \sqrt{3} < 2 + \sqrt{x+2} < 2 + \sqrt{5}.
• Bounding $\sqrt{x+2} + 4$: Similarly, since \sqrt{3} < \sqrt{x+2} < \sqrt{5}, then \sqrt{3} + 4 < \sqrt{x+2} + 4 < \sqrt{5} + 4.

Using these bounds, we can get a lower bound for the denominator:

• $2(\sqrt{x+2}+4)(2 + \sqrt{x+2}) > 2(\sqrt{3}+4)(2+\sqrt{3}) $

So, we have:

\frac{|x - 2|}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})} < \frac{|x - 2|}{2(\sqrt{3}+4)(2+\sqrt{3})}

Now, we want to make sure that \frac{|x - 2|}{2(\sqrt{3}+4)(2+\sqrt{3})} < \epsilon. To do this, we can set:

$$\delta = \min\{1, 2\epsilon(\sqrt{3}+4)(2+\sqrt{3})\}$$

This is our $\delta$. We choose the minimum of $1$ and $2\epsilon(\sqrt{3}+4)(2+\sqrt{3})$ because we initially assumed $\delta \le 1$. This ensures that our choice of $\delta$ satisfies both the initial assumption and the requirement to be less than $\epsilon$.

The Formal Proof: Putting it All Together

Let's write out the formal proof. Remember, the goal is to show that for any \epsilon > 0, we can find a \delta > 0 that satisfies the $\epsilon-\delta$ definition.

Proof:

1. Let \epsilon > 0 be given. We want to find a \delta > 0 such that if 0 < |x - 2| < \delta, then |\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}| < \epsilon.

2. Choose $\delta$: Let's choose $\delta = \min\{1, 2\epsilon(\sqrt{3}+4)(2+\sqrt{3})\}$.

3. Assume 0 < |x - 2| < \delta: This means that 0 < |x - 2| < 1 and 0 < |x - 2| < 2\epsilon(\sqrt{3}+4)(2+\sqrt{3}).

4. Manipulate the expression:

   • We already showed that $|\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}| = \frac{|x - 2|}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})}$.

   • Since we assumed |x-2| < 1, then 1 < x < 3. Therefore, 3 < x+2 < 5, which leads to \sqrt{3} < \sqrt{x+2} < \sqrt{5}.

   • This gives us 2+\sqrt{3} < 2 + \sqrt{x+2} and \sqrt{x+2} + 4 > \sqrt{3} + 4. The main idea here is to find the lower bound of the denominator. Thus we can get:

     \frac{|x - 2|}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})} < \frac{|x - 2|}{2(\sqrt{3}+4)(2+\sqrt{3})}

5. Apply the chosen $\delta$: Since 0 < |x - 2| < \delta, we have:

   \frac{|x - 2|}{2(\sqrt{x+2}+4)(2 + \sqrt{x+2})} < \frac{|x - 2|}{2(\sqrt{3}+4)(2+\sqrt{3})} < \frac{\delta}{2(\sqrt{3}+4)(2+\sqrt{3})}

   • We chose $\delta$ such that $\delta \le 2\epsilon(\sqrt{3}+4)(2+\sqrt{3})$. Thus we get

     $$\frac{\delta}{2(\sqrt{3}+4)(2+\sqrt{3})} \le \frac{2\epsilon(\sqrt{3}+4)(2+\sqrt{3})}{2(\sqrt{3}+4)(2+\sqrt{3})} = \epsilon$$

6. Conclusion: Therefore, |\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}| < \epsilon. This completes the proof. We have shown that for any \epsilon > 0, we can find a \delta > 0 such that if 0 < |x - 2| < \delta, then |\frac{3}{\sqrt{x+2}+4} - \frac{1}{2}| < \epsilon. So, the limit exists and is indeed $\frac{1}{2}$.

Final Thoughts

And there you have it, guys! We successfully proved the limit using the $\epsilon-\delta$ definition. It takes a bit of practice to get used to, but with each problem you solve, you'll become more confident. Remember to break down the problem step-by-step, manipulate the expressions, and always try to relate everything back to $|x - a|$. Keep practicing, and you'll master this fundamental concept in no time. Good luck and happy calculating!