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Chapter 9: Problem 746

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

Short Answer

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

Step by step solution

01

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
02

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
03

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
04

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
05

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
06

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
07

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).

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Most popular questions from this chapter

If \(\mathrm{k}^{\text {th }}\) term \(\mathrm{t}_{\mathrm{k}}\), of the series is formulated as \(\mathrm{t}_{\mathrm{k}}=\left[\mathrm{k} /\left(1+\mathrm{k}^{2}+\mathrm{k}^{4}\right)\right]\) then \(\lim _{\mathrm{n} \rightarrow \infty}^{\mathrm{n}} \sum_{\mathrm{k}=1} \mathrm{t}_{\mathrm{k}}=\ldots \ldots\) is: (a) \(0.25\) (b) \(0.50\) (c) 1 (d) Limit of the series does not exists

\(\lim _{\mathrm{x} \rightarrow \infty}\left[\left(\mathrm{x}^{2}+7 \mathrm{x}+2013\right) / \mathrm{x}^{2}\right]^{7 \mathrm{x}}=?\) (a) \(\mathrm{e}^{7}\) (b) \(\mathrm{e}^{14}\) (c) \(\mathrm{e}^{21}\) (d) \(\mathrm{e}^{49}\)

If \(\mathrm{f}(\mathrm{x})=\mid \begin{array}{ll}(\operatorname{Sin} 2 \mathrm{x})^{(\tan ) 2(2 \mathrm{x}) ;} ; & \mathrm{x} \neq(\pi / 4) \\\ \mathrm{K} ; & \mathrm{x}=(\pi / 4)\end{array}\) is continuous \(\mathrm{x} \neq(\pi / 4)\) then the value of \(\mathrm{K}\) is: (a) \(\mathrm{e}^{(1 / 2)}\) (b) \(\mathrm{e}^{-(1 / 2)}\) (c) \(\mathrm{e}^{2}\) (d) \(\mathrm{e}^{-2}\)

\(\lim _{\mathrm{x} \rightarrow-(\pi / 4)}[\\{\sin \mathrm{x} \cdot \cos (5 \pi / 4)-\cos (7 \pi / 4) \cos \mathrm{x}\\} /(\pi+4 \mathrm{x})]\) \(=\) ? \(\begin{array}{lll}\text { (a) }-(1 / 3) & \text { (b) }(35 / 4) & \text { (c) }-(1 / 4)\end{array}\) (d) \(-(1 / 35)\)

\(\lim _{\mathrm{x} \rightarrow 0}\left[\left\\{\sqrt{\left.\left. \left.\left(5+\mathrm{x}^{5}\right)-\sqrt{(5-\mathrm{x}}^{5}\right)\right\\} / \mathrm{x}^{5}\right]}=?\right.\right.\) (a) 5 (b) 25 (c) \((\sqrt{5})^{-1}\) (d) \(4 \sqrt{5}\)
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