/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 110 As $g(x)= \begin{cases}(a-x) \se... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 110

As $g(x)= \begin{cases}(a-x) \sec \frac{\pi x}{2 a}, & xa\end{cases}$ $\begin{aligned} \lim _{x \rightarrow a} g(x) &=\lim _{x \rightarrow a} \frac{(a-x)}{\cos \left(\frac{\pi x}{2 a}\right)} \\ &=\lim _{b \rightarrow 0} \frac{h}{\sin \left(\frac{\pi}{2} \frac{h}{a}\right)} \Rightarrow \frac{2 a}{\pi} \\ \lim _{x \rightarrow a^{\prime}} g(x)=\lim _{h \rightarrow 0}\left(-\cot \left(\frac{\pi}{2}+\frac{\pi}{2 a} h\right) \operatorname{cosec}(h)\right) \\ &=\lim _{h \rightarrow 0}+\frac{\tan \left(\frac{\pi}{2 a} h\right)}{\sin h} \times \frac{h}{\frac{\pi}{2 a} h} \times \frac{\pi}{2 a}=\frac{\pi}{2 a} \end{aligned}$ Hence, D is correct.

Short Answer

Expert verified
In this exercise, we needed to find the limit of a piecewise function \(g(x)\) as \(x\) approaches \(a\). We considered the two cases, \(xa\), separately. For both cases, we used a substitution with \(h\) to simplify the expressions and then applied limit properties and L'Hopital's rule to find the limits. The final result is: $$\lim_{x \rightarrow a} g(x) = \begin{cases} \frac{2a}{\pi}, &xa \end{cases}$$

Step by step solution

01

Find the limit as x approaches a for x < a

First, we need to find the limit of the first case: \(\lim_{x \rightarrow a} (a - x)\sec\left(\frac{\pi x}{2a}\right)\). We can use the substitution \(h=a-x\) and rewrite the limit as follows: $$\lim_{h \rightarrow 0} \frac{h}{\cos\left(\frac{\pi}{2} \cdot \frac{h}{a}\right)}$$
02

Use properties of limits

Use the property of limits to rewrite the limit as: $$\lim_{h \rightarrow 0}\frac{h}{\sin\left(\frac{\pi}{2} - \frac{\pi h}{2a}\right)}$$
03

Apply L'Hopital's rule

L'Hopital's rule states that if we have a limit in the form \(\lim_{h \rightarrow 0} \frac{f(h)}{g(h)}\), where both \(f(h)\) and \(g(h)\) tends to \(0\) as \(h\) goes to \(0\), then the limit equals to \(\lim_{h \rightarrow 0} \frac{f'(h)}{g'(h)}\). So in this case, using L'Hopital's rule: $$\lim_{h \rightarrow 0} \frac{1}{-\frac{\pi}{2a}\cos\left(\frac{\pi}{2} - \frac{\pi h}{2a}\right)} = \frac{2a}{\pi}$$ Now let's move on to the second case (\(x>a\)).
04

Find the limit as x approaches a for x > a

Similarly, we will consider the second case of \(g(x)=-\cot\left(\frac{\pi x}{2 a}\right)\csc(x-a)\). Use substitution \(h = x-a\). The limit becomes: $$\lim_{h \rightarrow 0} -\cot\left(\frac{\pi}{2} + \frac{\pi h}{2 a}\right)\operatorname{cosec}(h)$$
05

Rewrite in terms of tangent and sine functions

Rewrite the limit as the following: $$\lim_{h \rightarrow 0} \frac{\tan\left(\frac{\pi h}{2a}\right)}{\sin(h)} \times \frac{h}{\frac{\pi}{2a} h} \times \frac{\pi}{2a}$$
06

Apply limit properties

Now, apply limit properties to solve the given expression: $$\lim_{h \rightarrow 0} \left[\frac{\tan\left(\frac{\pi h}{2a}\right)}{\frac{\pi h}{2a}} \times \frac{h}{\sin(h)}\right] \times \frac{\pi}{2a}$$ Since \(\lim_{h \rightarrow 0} \frac{\tan(\pi h/2a)}{\pi h/2a} = 1\) and \(\lim_{h \rightarrow 0} \frac{h}{\sin(h)} = 1\), we have: $$\lim_{h \rightarrow 0}\left[-\cot\left(\frac{\pi}{2}+\frac{\pi}{2a}h\right)\operatorname{cosec}(h)\right] = \frac{\pi}{2a}$$
07

Conclusion

The limit of the piecewise function \(g(x)\) as \(x\) approaches \(a\) is: $$\lim_{x \rightarrow a} g(x) = \begin{cases} \frac{2a}{\pi}, &xa \end{cases}$$ Hence, the answer is correct.

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Most popular questions from this chapter

$\mathrm{f}(\mathrm{x})=\lim _{\mathrm{n} \rightarrow \infty} \frac{(2 \sin \mathrm{x})^{2 \mathrm{n}}}{3^{\mathrm{n}}-(2 \cos \mathrm{x})^{2 \mathrm{n}}}\( \)\Rightarrow \lim _{n \rightarrow \infty} \frac{(2 \sin x)^{2 n}}{1-\left(\frac{2}{\sqrt{3}} \cos x\right)^{2 n}}$ \(\mathrm{f}(\mathrm{x})\) is discontinuous whenever \(2 \sin x=\pm 1 \& \frac{2}{\sqrt{3}} \cos x=\pm 1\) \(\Rightarrow \mathrm{x}=\mathrm{n} \pi \pm \frac{\pi}{6}\)

$$ \lim _{x \rightarrow 1^{\prime}} F(x)=\lim _{x \rightarrow 1^{*}} \operatorname{sgn}(x-1) \cot ^{-1}[x-1] $$ $$ \begin{aligned} &=\frac{\pi}{2} \\ \lim _{x \rightarrow 1^{-}} F(x) &=\lim _{\left.x \rightarrow\right|^{-}} \operatorname{sgn}(x-1) \cot ^{-1}[x-1] \\ &=-\frac{3 \pi}{4} \end{aligned} $$

\(\lim _{x \rightarrow 0} \frac{2-\sqrt[4]{x^{2}+16}}{\cos 2 x-1}\) $\lim _{x \rightarrow 0} \frac{4-\sqrt[2]{x^{2}+16}}{\left(2+\sqrt[4]{x^{2}+16}\right)(\cos 2 x-1)}$ $\lim _{x \rightarrow 0} \frac{-x^{2}}{\left(2+\sqrt[4]{x^{2}+16}\right)\left(4+\sqrt[2]{x^{2}+16}\right)(\cos 2 x-1)}$ \(=\frac{1}{64}\)

\(\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\tan k x}{x}=k\) $\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left(3+2 k^{2}\right)=2 k^{2}$ For continuity, \(\mathrm{k}=2 \mathrm{k}^{2}\) \(\Rightarrow \mathrm{K}=0\) or \(\frac{1}{2}\) Hence, \(\mathrm{C}\) is correct.

For \(g(x)\) to be continuous, then \(\mathrm{a}>\) range of \(\mathrm{f}(\mathrm{x})\) \(\Rightarrow \mathrm{a}>\sqrt{26}\) least positive integral value of \(a=3\) Hence, B is correct.
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