Epsilon-Delta Proof: Limit Of (2x+5)/(√(x+1)+6) As X->3

Hey guys! Today, we're diving deep into the world of real analysis to tackle a classic problem: proving a limit using the rigorous epsilon-delta definition. Specifically, we want to show that

$$\lim_{x\to 3} \frac{2x+5}{\sqrt{x+1}+6} = \frac{11}{8}$$

Before we get started, let's break down what this actually means. The epsilon-delta definition is a way to make the intuitive idea of a limit precise. It basically says: "We can make the function's output as close as we want to the limit (11/8 in this case) by making the input x close enough to 3." "Close enough" is quantified by $\epsilon$ (how close the output is) and $\delta$ (how close the input x is to 3). Let's get into the details.

Understanding the Epsilon-Delta Definition

The $\epsilon$-$\delta$ definition of a limit states that for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - c| < \delta$, then $|f(x) - L| < \epsilon$. In our case, $c = 3$, $f(x) = \frac{2x+5}{\sqrt{x+1}+6}$, and $L = \frac{11}{8}$.

So, our goal is to show that for any $\epsilon > 0$, we can find a $\delta > 0$ such that if $0 < |x - 3| < \delta$, then

$$\left|\frac{2x+5}{\sqrt{x+1}+6} - \frac{11}{8}\right| < \epsilon$$

The Strategy

The general strategy is to start with the expression $|f(x) - L|$ and manipulate it algebraically until we can relate it to $|x - c|$, which in our case is $|x - 3|$. Then, we can choose a $\delta$ that depends on $\epsilon$ to make the inequality $|f(x) - L| < \epsilon$ hold.

It's like saying, "Okay, you want the output to be within $\epsilon$ of 11/8? No problem! I'll just make sure x is within $\delta$ of 3, where $\delta$ is this specific value I'm going to calculate based on $\epsilon$."

The Proof

Let's start by manipulating the expression inside the absolute value:

$$\left|\frac{2x+5}{\sqrt{x+1}+6} - \frac{11}{8}\right| = \left|\frac{8(2x+5) - 11(\sqrt{x+1}+6)}{8(\sqrt{x+1}+6)}\right| = \left|\frac{16x + 40 - 11\sqrt{x+1} - 66}{8(\sqrt{x+1}+6)}\right| = \left|\frac{16x - 26 - 11\sqrt{x+1}}{8(\sqrt{x+1}+6)}\right|$$

Now, this looks a bit intimidating, but we can simplify it further. Notice that we want to relate this expression to $|x - 3|$. Let's try to factor something out that looks like $(x - 3)$. To do this, we need to do some algebraic trickery.

Multiply the numerator and denominator by the conjugate of $16x - 26 - 11\sqrt{x+1}$ which, regarding the square root term, is $16x - 26 + 11\sqrt{x+1}$.

$$\left|\frac{16x - 26 - 11\sqrt{x+1}}{8(\sqrt{x+1}+6)} \cdot \frac{16x - 26 + 11\sqrt{x+1}}{16x - 26 + 11\sqrt{x+1}}\right| = \left|\frac{(16x - 26)^2 - 121(x+1)}{8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})}\right|$$

Expanding the numerator, we get:

$$\left|\frac{256x^2 - 832x + 676 - 121x - 121}{8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})}\right| = \left|\frac{256x^2 - 953x + 555}{8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})}\right|$$

Now, let's see if we can factor $(x-3)$ from the numerator. After some factoring (or using synthetic division), we find that:

$$256x^2 - 953x + 555 = (x - 3)(256x - 185)$$

So, our expression becomes:

$$\left|\frac{(x - 3)(256x - 185)}{8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})}\right| = |x - 3| \cdot \left|\frac{256x - 185}{8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})}\right|$$

Bounding the Expression

Now we have $|x - 3|$ nicely factored out. We need to bound the remaining expression. Since we're interested in x close to 3, let's assume that $|x - 3| < 1$. This implies that $2 < x < 4$.

Let's analyze the denominator:

• $\sqrt{x+1}$ is between $\sqrt{3} \approx 1.73$ and $\sqrt{5} \approx 2.24$, so $\sqrt{x+1} + 6$ is between roughly 7.73 and 8.24.
• $16x - 26$ is between $16(2) - 26 = 6$ and $16(4) - 26 = 38$.
• $11\sqrt{x+1}$ is between roughly $11(1.73) \approx 19.03$ and $11(2.24) \approx 24.64$.
• So, $16x - 26 + 11\sqrt{x+1}$ is between roughly $6 + 19.03 = 25.03$ and $38 + 24.64 = 62.64$.

Thus, the denominator $8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})$ is bounded below by approximately $8(7.73)(25.03) \approx 1548$.

Now, let's look at the numerator $|256x - 185|$. Since $2 < x < 4$, we have:

• $256x$ is between $256(2) = 512$ and $256(4) = 1024$.
• So, $256x - 185$ is between $512 - 185 = 327$ and $1024 - 185 = 839$.
• Thus, $|256x - 185|$ is bounded above by 839.

Therefore,

$$\left|\frac{256x - 185}{8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})}\right| < \frac{839}{1548} < 1$$

(A rough overestimate to simplify calculations. A more precise bound would work too).

Choosing Delta

We want to find a $\delta$ such that if $0 < |x - 3| < \delta$, then

$$\left|\frac{2x+5}{\sqrt{x+1}+6} - \frac{11}{8}\right| = |x - 3| \cdot \left|\frac{256x - 185}{8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})}\right| < \epsilon$$

Since we know that $\left|\frac{256x - 185}{8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})}\right| < 1$ (from our assumption that $|x-3|<1$), we have:

$$|x - 3| \cdot \left|\frac{256x - 185}{8(\sqrt{x+1}+6)(16x - 26 + 11\sqrt{x+1})}\right| < |x - 3|$$

So, if we choose $\delta = \min(1, \epsilon)$, then whenever $0 < |x - 3| < \delta$, we have $|x-3| < \epsilon$. Also, because we chose the minimum, we also have $|x-3| < 1$, satisfying our initial assumption used to bound the expression. Therefore, we have:

$$\left|\frac{2x+5}{\sqrt{x+1}+6} - \frac{11}{8}\right| < |x - 3| < \epsilon$$

Conclusion

We have shown that for any $\epsilon > 0$, there exists a $\delta = \min(1, \epsilon)$ such that if $0 < |x - 3| < \delta$, then $\left|\frac{2x+5}{\sqrt{x+1}+6} - \frac{11}{8}\right| < \epsilon$. Therefore, by the $\epsilon$-$\delta$ definition of a limit, we have proven that

$$\lim_{x\to 3} \frac{2x+5}{\sqrt{x+1}+6} = \frac{11}{8}$$

Woohoo! That was a tough one, but we made it through. The key is to carefully manipulate the expression, relate it to $|x - c|$, and then find a suitable $\delta$ in terms of $\epsilon$. Keep practicing, and you'll become an epsilon-delta master in no time!