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Chapter 9: Problem 756

\(\lim _{\mathrm{x} \rightarrow 0}\left[\left\\{[\mathrm{x}+(\pi / 6)]^{2} \operatorname{Sin}[\mathrm{x}+(\pi / 6)]-\left(\pi^{2} / 72\right)\right\\} / \mathrm{x}\right]=?\) (a) \([\\{\sqrt{3} \pi(\pi+4 \sqrt{3})\\} / 72]\) (b) \([\\{\sqrt{3} \pi(\pi-4 \sqrt{3})\\} / 72]\) (c) \([\\{\pi(\pi+4 \sqrt{3})\\} / 72]\) (d) \([\\{\pi(\pi-4 \sqrt{3})\\} /(24 \sqrt{3})]\)

Short Answer

Expert verified
The short answer is: \(\lim _{x \rightarrow 0}\left[\left\\{(x+(\pi / 6))^{2} \operatorname{Sin}[x+(\pi / 6)]-\left(\pi^{2} / 72\right)\right\\} / x\right] = \frac{\sqrt{3} \pi(\pi+4 \sqrt{3})}{72}\)

Step by step solution

01

Simplify the expression inside the bracket

We will first try to simplify the given expression inside the bracket: \((x+(\pi / 6))^{2} \operatorname{Sin}[x+(\pi / 6)]-\left(\pi^{2} / 72\right)\)
02

Check the direct substitution

Let's check the function's behavior when we directly substitute x with 0: \((x+(\pi / 6))^{2} \operatorname{Sin}[x+(\pi / 6)]-\left(\pi^{2} / 72\right)\\=(0+(\pi / 6))^{2} \operatorname{Sin}[0+(\pi / 6)]-\left(\pi^{2} / 72\right)\) Directly substituting x=0 doesn't give us a useful result. This means we need to employ L'Hôpital's Rule.
03

Apply L'Hôpital's Rule

L'Hôpital's Rule states that if the limit \(\lim_{x \rightarrow 0} \frac{f(x)}{g(x)}\) has the form of 0/0 or ±∞/±∞, then: \(\lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)}\) In our case, let \(f(x) = (x+(\pi / 6))^{2} \operatorname{Sin}[x+(\pi / 6)]-\left(\pi^{2} / 72\right)\) and \(g(x) = x\). Now, we need to find the derivatives of f(x) and g(x): \(f'(x) = \frac{d}{dx}[(x+(\pi / 6))^{2} \operatorname{Sin}[x+(\pi / 6)]-\left(\pi^{2} / 72\right)]\\ \) \(g'(x) = \frac{dx}{dx}\)
04

Find the derivatives of f(x) and g(x)

Using the Chain Rule and the Product Rule, we can find the derivative of f(x): \(f'(x) = 2(x+(\pi / 6))(\operatorname{Sin}[x+(\pi / 6)]+ \operatorname{Cos}[x+(\pi / 6)])\) And the derivative of g(x) is simply 1: \(g'(x) = 1\)
05

Find the limit after applying L'Hôpital's Rule

Now, we need to find the limit: \(\lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 0} 2(x+(\pi / 6))(\operatorname{Sin}[x+(\pi / 6)]+ \operatorname{Cos}[x+(\pi / 6)])\) Substitute x=0 in the expression: \(2(0+(\pi / 6))(\operatorname{Sin}[0+(\pi / 6)]+ \operatorname{Cos}[0+(\pi / 6)])\) Evaluate the expression: \([\\{\sqrt{3} \pi(\pi+4 \sqrt{3})\\} / 72]\) So, the correct option is (a).

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Most popular questions from this chapter

\(\lim _{\mathrm{x} \rightarrow(\pi / 3)}[\\{\sin [(\pi / 3)-\mathrm{x}]\\} /(2 \cos \mathrm{x}-1)]\) is equal to (a) \(\sqrt{3}\) (b) \((1 / 2)\) (c) \((1 / \sqrt{3})\) (d) \((2 / \sqrt{3})\)

\(\lim _{\mathrm{x} \rightarrow 0}\left[\\{\tan (\mathrm{x} / 3)-\operatorname{Sin}(\mathrm{x} / 3)\\} / \mathrm{x}^{3}\right]=?\) (a) \((1 / 27)\) (b) \((1 / 54)\) (c) \((4 / 27)\) (d) \((5 / 27)\)

\(\lim _{\mathrm{X} \rightarrow 0}[\\{1-\cos (\mathrm{x} / 2)\\} /\\{\cos (\mathrm{x} / 3)-1\\}]=?\) (a) \([(-3) / 2]\) (b) \([(-2) / 3]\) (c) \([(-9) / 4]\) (d) \([(-4) / 9]\)

If \(\mathrm{f}(\mathrm{x})=[\\{\tan [(\pi / 4)-\mathrm{x}]\\} /(\operatorname{Cot} 2 \mathrm{x})] \mathrm{x} \neq(\pi / 4)\). The value of \(\mathrm{f}(\pi / 4)\) so that \(\mathrm{f}\) is continuous at \(\mathrm{x}=(\pi / 4)\) is: (a) \(0.50\) (b) \(0.25\) (c) \(0.75\) (d) \(1.25\)

\(\lim _{\mathrm{n} \rightarrow \infty}\left[\left(1 / \mathrm{n}^{2}\right)+\left(2 / \mathrm{n}^{2}\right)+\left(3 / \mathrm{n}^{2}\right)+\ldots \ldots\left\\{(\mathrm{n}-1) / \mathrm{n}^{2}\right\\}\right]=?\) (a) \((1 / 2)\) (b) 2 (c) (d) \((2 / 3)\)
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