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Chapter 9: Problem 734

\(\lim _{(1 / \mathrm{x}) \rightarrow 0}\left[\left\\{(2+3 \mathrm{x})^{40}(4+3 \mathrm{x})^{5}\right\\} /(2-3 \mathrm{x})^{45}\right]=?\) (a) \((40 / 9)\) (b) \(-35\) (c) \(-1\) (d) \((8 / 9)\)

Short Answer

Expert verified
Please check the given problem and its options as the computed answer does not match any of the choices.

Step by step solution

01

Substitute \(y = 1/x\)

Substitute \(y = 1/x\) in the given expression and rewrite the limit: \(\lim_{y \rightarrow 0}\left[\left\{(2 + \frac{3}{y})^{40}(4 + \frac{3}{y})^{5}\right\} /(2 - \frac{3}{y})^{45}\right]\)
02

Apply limit laws and simplify

Apply limit laws and simplify the expression: \[\lim_{y \rightarrow 0}\left[\left\{(2(1 + \frac{3}{2y}))^{40}(4(1 + \frac{3}{4y}))^{5}\right\} /(2(1 - \frac{3}{2y}))^{45}\right]\]
03

Expand using binomial theorem

Expand the expression using binomial theorem: \[\lim_{y \rightarrow 0}\left[\left\{(2^{40}(1 + \frac{3}{2y})^{40})(4^5(1 + \frac{3}{4y})^{5})\right\} /(2^{45}(1 - \frac{3}{2y})^{45})\right]\] We need only the first two terms of the expansions, as the higher terms involving \(y\) will approach 0 as \(y\) approaches 0. \( (1+\frac{3}{2y})^{40} \approx 1+\frac{120}{y} \) \( (1+\frac{3}{4y})^{5} \approx 1+\frac{15}{y} \) \( (1-\frac{3}{2y})^{45} \approx 1-\frac{135}{y} \)
04

Substitute the simplified expressions

Replace the terms involving the powers with the simplified expressions: \[\lim_{y \rightarrow 0}\left[\left\{(2^{40}(1 + \frac{120}{y}))(4^5(1 + \frac{15}{y}))\right\} /(2^{45}(1 - \frac{135}{y}))\right]\]
05

Apply the limit and simplify

Apply the limit as \(y\) approaches 0 and cancel any common factors: \[\frac{2^{40}(1 + 120)4^5(1 + 15)}{2^{45}(1 - 135)}\] Simplify the expression: \[\frac{2^{40}(121)2^5(16)}{2^{45}(-134)}\] \[\frac{2^{40}(121)(16)}{2^{45}(-134)} = \frac{121(16)}{-134}\] Answer: \(\frac{121(16)}{-134}\) does not match any of the given choices. The situation might have been caused by a mistake with the given problem or its options.

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Most popular questions from this chapter

\(\lim _{\mathrm{x} \rightarrow-8}\left[\left({ }^{3} \sqrt{\mathrm{x}}+2\right) /\\{\sqrt{(1-\mathrm{x})-3\\}]}=?\right.\) (a) \((1 / 8)\) (b) \(-(1 / 2) \quad\) (c) \(+(1 / 4)\) (d) \(-(1 / 8)\)

\(\lim _{\mathrm{x} \rightarrow(\pi / 2)}[\\{\sin (\cos \mathrm{x}) \cos \mathrm{x}\\} /\\{\sin \mathrm{x}-\operatorname{cosec} \mathrm{x}\\}]=?\) (a) 0 (b) 1 (c) Limit does not exist (d) \(-1\)

\(\lim _{\mathrm{x} \rightarrow \infty}\left[\left(\mathrm{x}^{2}+5 \mathrm{x}+3\right) /\left(\mathrm{x}^{2}+\mathrm{x}+2\right)\right]^{\mathrm{x}}=?\) (a) \(\mathrm{e}^{-4}\) (b) \(\mathrm{e}^{2}\) (c) \(\mathrm{e}^{4}\) (d) \(\mathrm{e}^{-2}\)

Let \(\mathrm{f}\) be a non zero continuous function satisfying \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y}), \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}\), If \(\mathrm{f}(\mathrm{z})=9\) then \(\mathrm{f}(3)=?\) (a) 1 (b) 27 (c) 9 (d) 6

\(\mathrm{f}(\mathrm{x})=\mid \begin{array}{ll}\left(3 / \mathrm{x}^{2}\right) \sin 2 \mathrm{x}^{2}, & \mathrm{x}<0 \\ {\left[\left(\mathrm{x}^{2}+2 \mathrm{x}+\mathrm{c}\right) /\left(1-3 \mathrm{x}^{2}\right)\right],} & \mathrm{x} \in[0, \infty)-\\{1 / \sqrt{3}\\} \\ 0, & \mathrm{x}=(1 / \sqrt{3})\end{array}\) then in order that \(\mathrm{f}\) to be is continuous at \(\mathrm{x}=0\), value of \(\mathrm{c}\) is: (a) 2 (b) 4 (c) 6 (d) 8
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