Evaluate The Limit Of (x+6)tan(2x-6) / (x^2 - X - 6) As X Approaches 3

In the realm of calculus, evaluating limits is a fundamental concept that forms the bedrock of understanding continuity, derivatives, and integrals. This article delves into the process of evaluating a specific limit problem, providing a detailed step-by-step solution and highlighting key techniques. We will explore the limit of the function ${ \frac{(x+6) \tan(2x-6)}{x^2 - x - 6} }$ as x approaches 3. This problem encompasses several important concepts, including trigonometric functions, factorization, and the application of limit laws.

Understanding the Problem

Before diving into the solution, it's crucial to understand the problem statement. We are asked to find the limit of the given function as x gets arbitrarily close to 3. This means we need to determine the value the function approaches as x gets closer and closer to 3, without actually reaching 3. The function involves a rational expression with a tangent function in the numerator and a quadratic expression in the denominator. This suggests that we might need to employ algebraic manipulation, trigonometric identities, and limit laws to arrive at the solution.

The function we are dealing with is: ${ f(x) = \frac{(x+6) \tan(2x-6)}{x^2 - x - 6} }$

Our goal is to find: ${ \lim_{x \to 3} \frac{(x+6) \tan(2x-6)}{x^2 - x - 6} }$

Step 1: Initial Evaluation and Indeterminate Forms

The first step in evaluating any limit is to attempt direct substitution. This involves plugging in the value that x approaches (in this case, 3) into the function and see if we obtain a finite value. If we do, then that value is the limit. However, if we encounter an indeterminate form such as 0/0 or ∞/∞, we need to employ other techniques to evaluate the limit.

Let's substitute x = 3 into the function: ${ \frac{(3+6) \tan(2(3)-6)}{3^2 - 3 - 6} = \frac{9 \tan(0)}{9 - 3 - 6} = \frac{9 \cdot 0}{0} = \frac{0}{0} }$

As we can see, direct substitution results in the indeterminate form 0/0. This means we cannot directly determine the limit by substitution and need to proceed with algebraic manipulation.

Step 2: Factorization and Simplification

The presence of the indeterminate form 0/0 often suggests that there might be a common factor in the numerator and denominator that can be canceled out. In this case, we can factor the quadratic expression in the denominator: ${ x^2 - x - 6 = (x - 3)(x + 2) }$

Now, our function becomes: ${ f(x) = \frac{(x+6) \tan(2x-6)}{(x - 3)(x + 2)} }$

We still have a problem because substituting x=3 yields ${\tan(0)}$ in the numerator and ${(3-3)}$ in the denominator, so we need to manipulate the expression further.

Step 3: Trigonometric Manipulation and the Small Angle Approximation

To handle the tangent function, we can use the small angle approximation, which states that for small angles θ (in radians), tan(θ) ≈ θ. This approximation is particularly useful when dealing with limits involving trigonometric functions.

In our case, as x approaches 3, the argument of the tangent function, 2x - 6, approaches 0. Therefore, we can apply the small angle approximation: ${ \tan(2x - 6) \approx 2x - 6 }$

Substituting this approximation into our function, we get: ${ f(x) \approx \frac{(x+6)(2x-6)}{(x - 3)(x + 2)} }$

We can further simplify the numerator by factoring out a 2 from (2x - 6): ${ f(x) \approx \frac{2(x+6)(x-3)}{(x - 3)(x + 2)} }$

Now, we can cancel out the common factor of (x - 3) from the numerator and denominator, provided that x ≠ 3: ${ f(x) \approx \frac{2(x+6)}{x + 2} }$

Step 4: Evaluating the Limit After Simplification

After simplifying the function, we can now attempt direct substitution again. Let's substitute x = 3 into the simplified expression: ${ \lim_{x \to 3} \frac{2(x+6)}{x + 2} = \frac{2(3+6)}{3 + 2} = \frac{2(9)}{5} = \frac{18}{5} }$

Therefore, the limit of the function as x approaches 3 is 18/5.

Step 5: Rigorous Justification (Optional)

While the above steps provide a clear and intuitive solution, it's important to understand the underlying mathematical rigor. The small angle approximation, tan(θ) ≈ θ, is a consequence of the limit: ${ \lim_{θ \to 0} \frac{\tan(θ)}{θ} = 1 }$

To justify our use of the approximation, we can rewrite our original limit as follows: ${ \lim_{x \to 3} \frac{(x+6) \tan(2x-6)}{x^2 - x - 6} = \lim_{x \to 3} \frac{(x+6) \tan(2x-6)}{(x - 3)(x + 2)} }$

Now, multiply and divide by (2x - 6): ${ \lim_{x \to 3} \frac{(x+6) \tan(2x-6)}{(x - 3)(x + 2)} \cdot \frac{2x-6}{2x-6} = \lim_{x \to 3} \frac{(x+6) \tan(2x-6)}{(2x-6)} \cdot \frac{2x-6}{(x - 3)(x + 2)} }$

We can rewrite 2x - 6 as 2(x - 3): ${ \lim_{x \to 3} \frac{(x+6) \tan(2x-6)}{(2x-6)} \cdot \frac{2(x-3)}{(x - 3)(x + 2)} }$

Now, we can cancel out the (x - 3) terms and apply the limit: ${ \lim_{x \to 3} \frac{(x+6) \tan(2x-6)}{2(x-3)} \cdot \frac{2}{x + 2} }$

Let θ = 2x - 6. As x approaches 3, θ approaches 0. We can rewrite the limit as: ${ \lim_{x \to 3} (x+6) \cdot \lim_{x \to 3} \frac{\tan(2x-6)}{2x-6} \cdot \lim_{x \to 3} \frac{2}{x + 2} }$

Which becomes: ${ \lim_{x \to 3} (x+6) \cdot \lim_{θ \to 0} \frac{\tan(θ)}{θ} \cdot \lim_{x \to 3} \frac{2}{x + 2} }$

Using the fact that ${ \lim_{θ \to 0} \frac{\tan(θ)}{θ} = 1 }$, we get: ${ (3+6) \cdot 1 \cdot \frac{2}{3 + 2} = 9 \cdot \frac{2}{5} = \frac{18}{5} }$

This rigorous justification confirms our earlier result obtained using the small angle approximation.

Conclusion

In this article, we have demonstrated a comprehensive approach to evaluating the limit of the function ${ \frac{(x+6) \tan(2x-6)}{x^2 - x - 6} }$ as x approaches 3. We started by attempting direct substitution, which led to the indeterminate form 0/0. We then employed algebraic manipulation, factorization, and the small angle approximation to simplify the function. Finally, we evaluated the limit of the simplified function, arriving at the answer 18/5. We also provided a rigorous justification using the fundamental limit ${ \lim_{θ \to 0} \frac{\tan(θ)}{θ} = 1 }$. This problem highlights the importance of understanding various techniques for evaluating limits, including algebraic manipulation, trigonometric identities, and limit laws. By mastering these techniques, you can confidently tackle a wide range of limit problems in calculus and beyond.

Therefore, the correct answer is (D) 18/5.

This detailed solution not only provides the answer but also explains the reasoning and techniques involved, making it a valuable resource for students learning about limits in calculus.