Evaluating Limits A Comprehensive Guide To Limx→0 (2-x-2√(2-2x))
In the realm of calculus, limits serve as a fundamental concept that underpins many advanced topics. Understanding how to evaluate limits is crucial for grasping the behavior of functions as they approach specific points. This article delves into the evaluation of a particular limit: limx→0 (2-x-2√(2-2x)). We will explore the various techniques and steps involved in solving this limit, providing a comprehensive understanding of the process. We will begin by examining the initial form of the expression and identifying why direct substitution is not immediately applicable. Then, we will employ algebraic manipulation to transform the expression into a more manageable form, ultimately leading to the evaluation of the limit. This exploration will not only provide the solution to the specific limit but also enhance your problem-solving skills in calculus.
When confronted with a limit problem, the first step is often to attempt direct substitution. This involves replacing the variable (in this case, x) with the value it approaches (here, 0). Let's apply this to our limit:
limx→0 (2-x-2√(2-2x))
Substituting x = 0, we get:
2 - 0 - 2√(2 - 2(0)) = 2 - 2√2
This yields 2 - 2√2, which is a real number. However, this direct substitution doesn't lead to an indeterminate form, but it does show us that the function is continuous at x=0. Let's further analyze the expression to ensure we are not missing any nuances.
In many limit problems, direct substitution leads to indeterminate forms such as 0/0 or ∞/∞. These forms indicate that further manipulation is required to evaluate the limit. In our case, the initial substitution doesn't result in an indeterminate form, but to rigorously evaluate the limit, we need to employ algebraic techniques to simplify the expression and remove any potential ambiguities.
The key to solving this limit lies in algebraic manipulation. Specifically, we will employ a technique called rationalization. Rationalization involves eliminating radicals from an expression. In our case, we have a square root term, √(2-2x), which complicates the direct evaluation of the limit. To rationalize the expression, we multiply the numerator and denominator by the conjugate of the expression.
Our expression is (2-x-2√(2-2x)). To rationalize this, we treat (2-x) as one term and -2√(2-2x) as another term. The conjugate is then (2-x+2√(2-2x)). Multiplying the original expression by its conjugate divided by itself (which is equivalent to multiplying by 1) gives us:
[(2-x-2√(2-2x)) * (2-x+2√(2-2x))] / (2-x+2√(2-2x))
Now, we perform the multiplication in the numerator:
(2-x)^2 - (2√(2-2x))^2
Expanding this, we get:
(4 - 4x + x^2) - 4(2 - 2x) = 4 - 4x + x^2 - 8 + 8x = x^2 + 4x - 4
So, our expression becomes:
(x^2 + 4x - 4) / (2-x+2√(2-2x))
Now, let's rewrite the entire limit with this rationalized form:
limx→0 (x^2 + 4x - 4) / (2-x+2√(2-2x))
After rationalizing the expression, we have:
limx→0 (x^2 + 4x - 4) / (2-x+2√(2-2x))
Now, we can attempt direct substitution again. Substituting x = 0, we get:
(0^2 + 4(0) - 4) / (2 - 0 + 2√(2 - 2(0)))
This simplifies to:
(-4) / (2 + 2√2)
This is a real number and does not represent an indeterminate form. However, we should further simplify this expression to its simplest form. We can factor out a 2 from the denominator:
-4 / (2(1 + √2)) = -2 / (1 + √2)
To further rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is (1 - √2):
[-2 * (1 - √2)] / [(1 + √2) * (1 - √2)]
This simplifies to:
(-2 + 2√2) / (1 - 2) = (-2 + 2√2) / (-1) = 2 - 2√2
However, there seems to be a mistake in our calculations. Let's backtrack and identify the error. When we expanded (2-x)^2 - (2√(2-2x))^2, we should have:
(4 - 4x + x^2) - 4(2 - 2x) = 4 - 4x + x^2 - 8 + 8x = x^2 + 4x - 4
This seems correct. The error likely occurred in the initial rationalization step. Let's revisit that. We had:
[(2-x-2√(2-2x)) * (2-x+2√(2-2x))] / (2-x+2√(2-2x))
Expanding the numerator gives:
(2-x)^2 - (2√(2-2x))^2 = (4 - 4x + x^2) - 4(2 - 2x) = 4 - 4x + x^2 - 8 + 8x = x^2 + 4x - 4
Now, the original expression was (2-x-2√(2-2x)). We need to find a way to isolate x to simplify the limit. Let's try a different approach. Instead of rationalizing directly, let's try a substitution. Let u = √(2-2x). Then u^2 = 2-2x, and 2x = 2-u^2, so x = 1 - u^2/2. As x→0, u→√2. Our limit becomes:
limu→√2 [2 - (1 - u^2/2) - 2u]
This simplifies to:
limu→√2 [2 - 1 + u^2/2 - 2u] = limu→√2 [1 + u^2/2 - 2u]
This substitution doesn't seem to simplify the problem significantly. Let's go back to our rationalized form:
limx→0 (x^2 + 4x - 4) / (2-x+2√(2-2x))
Substituting x = 0, we get:
(-4) / (2 + 2√2) = -2 / (1 + √2)
Rationalizing the denominator:
[-2(1 - √2)] / [(1 + √2)(1 - √2)] = [-2 + 2√2] / (1 - 2) = [-2 + 2√2] / (-1) = 2 - 2√2
This is still not one of the provided options. We must have made an error in our simplification or the original question might have a typo. Let's consider the possibility of a typo in the original expression. Suppose the expression was intended to be:
limx→0 (2-x-2√(2-x))
In this case, let's rationalize:
[(2-x-2√(2-x)) * (2-x+2√(2-x))] / (2-x+2√(2-x))
Expanding the numerator:
(2-x)^2 - 4(2-x) = 4 - 4x + x^2 - 8 + 4x = x^2 - 4
So the limit becomes:
limx→0 (x^2 - 4) / (2-x+2√(2-x))
Substituting x = 0, we get:
(-4) / (2 + 2√2) = -2 / (1 + √2)
This leads to the same result as before. Let's try L'Hôpital's Rule on the original expression:
limx→0 (2-x-2√(2-2x))
The derivative of the numerator is:
-1 - 2 * (1/2)(2-2x)^(-1/2) * (-2) = -1 + 2(2-2x)^(-1/2)
Plugging in x = 0:
-1 + 2(2)^(-1/2) = -1 + 2/√2 = -1 + √2
Since we are getting a non-zero derivative, let's consider the original expression and see if we can apply L'Hôpital's Rule directly.
Rewrite the expression as:
limx→0 [2-x-2√(2-2x)] / 1
The derivative of the numerator with respect to x is:
-1 - 2 * (1/2)(2-2x)^(-1/2) * (-2) = -1 + 2(2-2x)^(-1/2)
As x→0, the derivative becomes:
-1 + 2(2)^(-1/2) = -1 + 2/√2 = -1 + √2 ≈ 0.414
Since the denominator's derivative is 0, we cannot directly apply L'Hôpital's Rule in this form. Let's revisit the rationalization and see if we can correct our previous mistake. We have:
limx→0 (2-x-2√(2-2x))
Multiplying by the conjugate (2-x+2√(2-2x)):
limx→0 [(2-x)^2 - 4(2-2x)] / [2-x+2√(2-2x)]
limx→0 [4 - 4x + x^2 - 8 + 8x] / [2-x+2√(2-2x)]
limx→0 [x^2 + 4x - 4] / [2-x+2√(2-2x)]
Substituting x = 0:
-4 / (2 + 2√2) = -2 / (1 + √2)
Rationalizing the denominator:
-2(1 - √2) / (1 - 2) = -2(1 - √2) / (-1) = 2 - 2√2
This is still not among the options. Let's re-examine L'Hôpital's Rule. We have the original limit:
limx→0 (2-x-2√(2-2x))
This is not in the indeterminate form 0/0 or ∞/∞, so we cannot directly apply L'Hôpital's Rule. However, if the limit were approaching an indeterminate form, we would have taken the derivative of the numerator and denominator separately. The derivative of the numerator is:
d/dx [2 - x - 2√(2 - 2x)] = -1 - 2 * (1/2)(2 - 2x)^(-1/2) * (-2) = -1 + 2/√(2 - 2x)
As x approaches 0, this becomes:
-1 + 2/√2 = -1 + √2
The derivative of the denominator (which is 1) is 0. Therefore, we can't use L'Hôpital's Rule in this situation. The only way we would have gotten to L'Hôpital's Rule is if the original question was structured as a fraction yielding the indeterminate form 0/0. Since this doesn't seem to be the case, and none of the simpler evaluations yield a clean integer, the most probable answer may be 0 due to potential rounding in a numerical analysis context, but without further context or correction to the limit expression, we cannot confidently select an option.
Conclusion
After careful analysis and application of various techniques, including rationalization and an attempt to apply L'Hôpital's Rule, we found that the limit limx→0 (2-x-2√(2-2x)) does not directly correspond to any of the provided options (-2, 0, 1, Does not exist). The result we consistently arrived at was 2 - 2√2, which is approximately -0.828. This discrepancy suggests a potential issue with the problem statement or the provided answer choices. It is crucial to double-check the original question for any typos or errors. In cases where a limit does not match the expected form, it often indicates a need for reevaluation or a possible error in the problem setup.
