Question

\(\left.\lim _{\mathrm{X} \rightarrow 2}\left[\left(2^{\mathrm{x}}+2^{3-\mathrm{x}}-6\right) /\left\\{\sqrt{(} 2^{-\mathrm{x}}\right)-2^{1-\mathrm{x}}\right\\}\right]=?\) (a) \(-12\) (b) 8 (c) \(-8\) (d) 6

Short answer

The limit of the given function as x approaches 2 is \(-8\), which corresponds to option (c).

Step-by-step solution

Identify the function

We are given the function in the exercise: \(\frac{2^x + 2^{3-x} - 6}{\sqrt{2^{-x}} - 2^{1-x}}\) and asked to find the limit as x approaches 2. Step 2: Calculate the limit as x approaches 2

Calculate the limit as x approaches 2

We will now calculate the limit of the function as x approaches 2. Notice that when we substitute x with 2, the denominator becomes 0. So we will have to solve the given function by applying the L'Hôpital's Rule. Step 3: Apply L'Hôpital's Rule

Apply L'Hôpital's Rule

To apply L'Hôpital's Rule, we need to take the derivative of both the numerator and the denominator. We will then be able to evaluate the limit by substituting x = 2 again. Step 4: Derivative of the numerator

Derivative of the numerator

Derivative of \(2^x + 2^{3-x} - 6\) (using chain rule): \(D=\frac{d}{dx}(2^x) + \frac{d}{dx}(2^{3-x}) - \frac{d}{dx}(6)\) \(D=2^x\ln2 - 2^{3-x}\ln2\) Step 5: Derivative of the denominator

Derivative of the denominator

Derivative of \(\sqrt{2^{-x}} - 2^{1-x}\) (using chain rule): \(D= -\frac{1}{2}(2^{-x})^{-\frac{1}{2}}(-\ln2)(2^{-x}) - 2^{1-x}\ln2\) \(D= -\sqrt{2^{-x}}\ln2(2^{-x}+2^{-x})\) Step 6: Apply L'Hôpital's Rule to the derivative

Apply L'Hôpital's Rule to the derivative

The L'Hôpital's Rule states that, if a function has the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) at x = c, then \(\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}\) Using the derivatives we found in Steps 4 and 5, we will now evaluate the limit at x = 2: \(\lim_{x\to 2}\frac{2^x\ln2 - 2^{3-x}\ln2}{-\sqrt{2^{-x}}\ln2(2^{-x}+2^{-x})} = \frac{2^2\ln2 - 2^{3-2}\ln2}{-\sqrt{2^{-2}}\ln2(2^{-2}+2^{-2})}\) Step 7: Solve and simplify

Solve and simplify

We can now further evaluate and simplify the expression from Step 6: \(\frac{4\ln2 - 2\ln2}{-\sqrt{\frac{1}{4}}\ln2(\frac{1}{4}+\frac{1}{4})} = \frac{2\ln2}{-\frac{1}{2}\ln2(\frac{1}{2})} = \frac{2\ln2}{-\frac{1}{4}\ln2}\) Dividing both numerator and denominator by \(\frac{1}{4}\ln2\), we get: \(\frac{2\ln2 \div \frac{1}{4}\ln2}{-\frac{1}{4}\ln2 \div \frac{1}{4}\ln2} = \frac{(2\div\frac{1}{4})(\ln2\div\ln2)}{-1}= - 8\) Answer: The limit of the given function as x approaches 2 is \(-8\), which corresponds to option (c).

Key concepts

Limit of a Function

The concept of the limit of a function is fundamental in calculus. When we discuss the limit of a function, we are essentially looking at what value a function approaches as the input—the variable—moves closer to a specific point. In mathematical notation, this is expressed as \( \lim_{x \to a} f(x) \), which asks what value \( f(x) \) approaches as \( x \) approaches \( a \). Understanding this concept allows us to analyze the behavior of functions near specific points, especially where the function might not be explicitly defined. It is crucial in determining points of discontinuity, behavior at infinity, or evaluating forms that seem undefined at first glance.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change of the function with respect to its variable. In simpler terms, it tells us how the function's value changes as the input changes. For instance, for a function \( f(x) \), its derivative is denoted \( f'(x) \) or \( \frac{df}{dx} \). The process involves utilizing various rules such as the power rule, product rule, and the chain rule to differentiate complex expressions. This process is essential when applying L'Hôpital's Rule, as it relies on taking derivatives of the numerator and denominator to resolve indeterminate forms.

Chain Rule

The chain rule is a technique used in differentiation when dealing with composite functions. A composite function is a function within another function, for example, \( f(g(x)) \). The chain rule states that if you have a function composed of two functions, you differentiate the outer function while keeping the inner function intact, and multiply by the derivative of the inner function. In formula terms, if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). This rule is particularly useful in complex problems where functions are nested within each other, such as when differentiating exponentials or roots that contain variable expressions.

Indeterminate Forms

Indeterminate forms arise in calculus when limits result in expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms do not provide immediate answers and require further analysis to resolve. L'Hôpital's Rule is a common method for dealing with these forms, allowing us to differentiate the numerator and denominator independently to bypass the indeterminacy. By doing this, we transform an unsolvable limit problem into a new one that can often be directly evaluated. Recognizing these forms is essential, as they tell us more work is needed to understand the true behavior of the function at the point under consideration.