Proving the Limit of sin(x)/(3x-5) as x Approaches π Using the ε-δ Definition

To prove the limit of (frac{sin(x)}{3x-5}) as (x) approaches (pi) using the ε-δ definition of limits, we first need to establish a few basic properties and continuity conditions. We will then proceed step-by-step through the rigorous proof process.

Conceptual Basis

The function (sin(x)) and the linear function (3x-5) are both continuous. Additionally, at (xpi), (3pi - 5 eq 0). This is crucial for our proof as it ensures the denominator is non-zero at the point of interest.

Theorem

Consider two continuous functions (f(x)) and (g(x)) such that (g(x) eq 0) in a neighborhood of (a). If (a eq 0), then the limit of (frac{f(x)}{g(x)}) as (x) approaches (a) is given by:

[lim_{x to a} frac{f(x)}{g(x)} frac{f(a)}{g(a)}]

To prove this, we use the ε-δ definition of limits. Let (epsilon > 0) be given. We must find a (delta > 0) such that for all (x) satisfying (0

[left|frac{f(x)}{g(x)} - frac{f(a)}{g(a)}right|

This can be rewritten as:

[left|frac{f(x)g(x) - f(g(x))}{g(x)g(a)}right|

Further, we can split this into two parts:

[left|frac{f(x)g(x) - f(a)g(x) f(a)g(x) - f(a)g(a)}{g(x)g(a)}right|

This can be simplified to:

[left|frac{f(x)g(x) - f(a)g(x) f(a)g(x) - f(a)g(a)}{g(x)g(a)}right| left|frac{g(x)(f(x) - f(a)) - f(a)(g(x) - g(a))}{g(x)g(a)}right|]

Thus, we have:

[left|frac{f(x)}{g(x)g(a)}(g(x) - g(a)) - frac{f(a)}{g(a)}(g(x) - g(a))right|]

Since (g(x)) and (f(x)) are continuous, there exists a (delta_1 > 0) and (delta_2 > 0) such that for all (x) satisfying (0

[left|frac{f(x)}{g(x)g(a)} - frac{f(a)}{g(a)}right| cdot |g(x) - g(a)|

and for all (x) satisfying (0

[left|frac{f(x)}{g(x)g(a)}right| cdot |g(x) - g(a)|

Let (delta min(delta_1, delta_2)). Then, for all (x) satisfying (0

[left|frac{f(x)}{g(x)} - frac{f(a)}{g(a)}right|

Hence, by the ε-δ definition, we have proven that (lim_{x to a} frac{f(x)}{g(x)} frac{f(a)}{g(a)}).

Applying the Theorem to sin(x)/(3x-5)

Now, let's apply the theorem to (f(x) sin(x)) and (g(x) 3x - 5), with (a pi).

Since (sin(x)) and (3x - 5) are continuous, and (3pi - 5 eq 0), the theorem is applicable. We need to show that (lim_{x to pi} frac{sin(x)}{3x - 5} frac{sin(pi)}{3pi - 5}). Since (sin(pi) 0), we have: [lim_{x to pi} frac{sin(x)}{3x - 5} 0].

To demonstrate this formally, let (epsilon > 0) be given. We need to find a (delta > 0) such that for all (x) satisfying (0

[left|frac{sin(x)}{3x - 5}right|

Given that (sin(x)) is continuous and bounded, there exists a (delta > 0) such that for all (x) satisfying (0

[|sin(x)|

Multiplying both sides by (left|frac{1}{3x - 5}right|), we get:

[left|frac{sin(x)}{3x - 5}right|

This proves that (lim_{x to pi} frac{sin(x)}{3x - 5} 0).

Conclusion

In conclusion, by leveraging the ε-δ definition and the properties of continuous functions, we have rigorously proven that the limit of (frac{sin(x)}{3x-5}) as (x) approaches (pi) is indeed 0.