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Chapter 9: Problem 723

\(\lim _{\mathrm{x} \rightarrow 2}\left[\left\\{\left({ }^{4} \sum_{\mathrm{i}=1} \mathrm{x}^{\mathrm{i}}\right)-30\right\\} /\left(\mathrm{x}^{3}-8\right)\right]=?\) (a) \((5 / 2)\) (b) \((8 / 2)\) (c) \((49 / 12)\) (d) \((43 / 12)\)

Short Answer

Expert verified
(c) \(\frac{49}{12}\)

Step by step solution

01

Evaluate the summation series

First, let's evaluate the summation series for \(i\) from 1 to 4: \[\sum_{i=1}^4 x^i = x^1 + x^2 + x^3 + x^4\]
02

Substitute the summation series in the given expression

Now, substitute the summation series back into the given expression: \[\frac{x^1 + x^2 + x^3 + x^4 - 30}{x^3 - 8}\]
03

Factor the denominator

We notice that the denominator is a difference of cubes. Let's factor it: \[\frac{x^1 + x^2 + x^3 + x^4 - 30}{x^3 - 8} = \frac{x^1 + x^2 + x^3 + x^4 - 30}{(x-2)(x^2+2x+4)}\]
04

Use the limit definition

Now that the expression is simplified, we can use the limit definition and substitute x = 2: \[\lim_{x\rightarrow 2}\frac{x^1 + x^2 + x^3 + x^4 - 30}{(x-2)(x^2+2x+4)} = \frac{2^1 + 2^2 + 2^3 + 2^4 - 30}{(2-2)(2^2+2\cdot 2+4)}\]
05

Simplify the expression and find the limit

Simplify the expression: \(= \frac{2 + 4 + 8 + 16 - 30}{(0)(4+4+4)} = \frac{0}{0}\) The limit is in the form \(\frac{0}{0}\), which indicates that we need to use l'Hopital's Rule or factor the numerator further to cancel the (x-2) term.
06

Factor the numerator to cancel (x-2) term

The common factor in the numerator is x, so we factor x out of the numerator: \[\frac{x(x^3 + x^2 + x^1 + x^0 - 30/x)}{(x-2)(x^2+2x+4)}\] By plugging in x = 2 in the updated expression, we get: \[\frac{2(2^3 + 2^2 + 2^1 + 1 - 30/2)}{(2-2)(2^2+2\cdot 2+4)}\] Simplify the expression: \(= \frac{2(8 + 4 + 2 + 1 - 15)}{(0)(4+4+4)} = \frac{0}{0}\) Still, the limit is in the form \(\frac{0}{0}\), so we have to apply l'Hopital's rule to find the limit.
07

Apply l'Hopital's Rule

Applying l'Hopital's Rule, we differentiate the numerator and the denominator with respect to x: Numerator derivative: \(1 + 2x + 3x^2 + 4x^3\) Denominator derivative: \((1)(x^2 + 2x + 4) + 2x(x-2)\) Now, we find the limit of the quotient of the derivatives as x approaches 2: \[\lim_{x\rightarrow 2}\frac{1 + 2x + 3x^2 + 4x^3}{(x^2 + 2x + 4) + 2x(x-2)}\] Plug in x = 2: \(= \frac{1 + (2)(2) + (3)(2)^2 + (4)(2)^3}{(2)^2 + (2)(2)+(4) + (2)(2)(2-2)}\) Simplify the expression: \(= \frac{1+4+12+32}{4+4+4} = \frac{49}{12}\) So, the limit is \(\frac{49}{12}\). Therefore, the correct option is (c) \(\frac{49}{12}\).

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Most popular questions from this chapter

The value of \(\mathrm{k}(\mathrm{k}>0)\) for which the function \(\mathrm{f}(\mathrm{x})=\left[\left(\mathrm{e}^{\mathrm{x}}-1\right)^{4} /\left\\{\operatorname{Sin}\left(\mathrm{x}^{2} / \mathrm{k}^{2}\right) \log \left[1+\left(\mathrm{x}^{2} / 2\right)\right]\right\\}\right]\) \(\mathrm{x} \neq 0, \mathrm{f}(0)=8\) may be continuous at \(\mathrm{x}=0\) is: (a) 1 (b) 2 (c) 4 (d) 3

\(\lim _{\mathrm{x} \rightarrow 0}\left[\left\\{(1+99 \mathrm{x})^{100}-(1+100 \mathrm{x})^{99}\right\\} / \mathrm{x}^{2}\right]=?\) (a) \(-4950\) (b) 4950 (c) 9950 (d) \(-9900\)

\(\lim _{(1 / x) \rightarrow 0} x \cdot \cos (\pi / 12 x) \sin (\pi / 12 x)=? \quad(x>0)\) (a) \((\pi / 12)\) (b) \((3 / \pi)\) (c) \((\pi / 3)\) (d) \((12 / \pi)\)

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) \(\mathrm{e}^{-2}\)

\(\lim _{x \rightarrow \sqrt{2}}\left[\left(x^{9}-3 x^{8}+x^{6}-9 x^{4}-4 x^{2}-16 x+84\right)\right.\) \(\left./\left(x^{5}-3 x^{4}-4 x+12\right)\right]=?\) (a) \(11+\sqrt{2}\) (b) \(11-\sqrt{2}\) (c) \(\sqrt{2}-11\) (b) \(11+2 \sqrt{2}\)
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