/*! 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 20 As \(f(b)=g(b) \&\) both are... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 20

As \(f(b)=g(b) \&\) both are Continuous function, then $h\left(b^{-}\right)=g\left(b^{+}\right) \& h\left(b^{+}\right)=f\left(b^{-}\right)$

Short Answer

Expert verified
Based on the step-by-step solution, the given conditions, \(h\left(b^{-}\right) = g\left(b^{+}\right)\) and \(h\left(b^{+}\right) = f\left(b^{-}\right)\), have been demonstrated to be true using the function \(h(x)\) and the continuity of \(f\) and \(g\). This confirms the validity of the conditions as stated in the exercise.

Step by step solution

01

Define \(h\) Function

Define a new function \(h\) as follows: $$h(x) = \begin{cases} f(x) & \text{if } x \leq b \\ g(x) & \text{if } x > b \end{cases}$$
02

Evaluate \(h\) at \(b\)

Since \(f\) and \(g\) are continuous and \(f(b) = g(b)\), it follows that \(h\) is continuous at \(b\). This means we can evaluate \(h\) at \(b\) using either \(f\) or \(g\). Let's use \(f\): $$h(b) = f(b) = g(b)$$
03

Determine \(h(b^-)\) and \(h(b^+)\)

Now that we have a value for \(h(b)\), we need to determine the one-sided limits of \(h\) at \(b\). Using the definition of \(h\), we know that \(h(x) = f(x)\) for \(x \leq b\) and \(h(x) = g(x)\) for \(x > b\). For \(h\left(b^{-}\right)\), we are interested in the limit of \(h\) as \(x\) approaches \(b\) from the left. Since we know that \(h(x) = f(x)\) for \(x \leq b\), this means that: $$ h\left(b^{-}\right) = f\left(b^{-}\right) $$ For \(h\left(b^{+}\right)\), we are interested in the limit of \(h\) as \(x\) approaches \(b\) from the right. Since we know that \(h(x) = g(x)\) for \(x > b\), this means that: $$ h\left(b^{+}\right) = g\left(b^{+}\right) $$
04

Verify the Given Conditions

We are given the following conditions to prove: 1. \(h\left(b^{-}\right) = g\left(b^{+}\right)\) 2. \(h\left(b^{+}\right) = f\left(b^{-}\right)\) Using our determined values for the one-sided limits of \(h\) at \(b\), we can verify these conditions: 1. \(f\left(b^{-}\right) = g\left(b^{+}\right)\) (True, by Step 3) 2. \(g\left(b^{+}\right) = f\left(b^{-}\right)\) (True, by Step 3) Both conditions are true, which concludes the exercise.

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Most popular questions from this chapter

$\prod_{n=1}^{n} \cos \left(\frac{x}{2^{n}}\right)=\frac{\sin x}{2^{n} \sin \left(\frac{x}{2^{n}}\right)}$ Taking log \& differentiate w.rt, \(x\). $\sum_{n=1}^{n} \frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{1}{2^{n}} \frac{\cos \left(x / 2^{n}\right)}{\sin \left(\frac{x}{2^{n}}\right)}-\cot x .$ now, $\lim _{n \rightarrow \infty} \sum_{n=1}^{n} \frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{1}{x}-\cot x$ $\lim _{x \rightarrow \frac{\pi}{2}} \lim _{n \rightarrow \infty} \sum_{n=1}^{n} \frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{2}{\pi}$

A) $f(x)= \begin{cases}\frac{1+a \cos x}{x^{2}}, & x<0 \\ b \tan \left(\frac{\pi}{[x+3]}\right), & x \geq 0\end{cases}$ \(\lim _{x \rightarrow 0^{-}} \frac{1+a \cos x}{x^{2}}\) Exists if \(\mathrm{a}=-1\). Then limit is \(\frac{1}{2}\). $\lim _{x \rightarrow 0^{-}} b+a \frac{\pi}{3}=\frac{1}{2} \Rightarrow b=\frac{1}{2 \sqrt{3}} \Rightarrow[a-2 b]=-2$ Hence, \(\mathrm{R}\) is correct. B) $\lim _{x \rightarrow-\frac{\pi}{2}}(-2 \sin x)=2$ \(\lim _{x \rightarrow-\frac{\pi^{*}}{2}}(a \sin x+b)=+b-a\) \(\lim _{x \rightarrow \frac{\pi}{2}}(\operatorname{asin} x+b)=a+b\) \(\lim _{x \rightarrow \frac{\pi}{2}} \cos x=0\) Hence \(\mathrm{P}, \mathrm{Q}\) is correct. C) $\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{3}{2}\right)^{\frac{\cos 3 x}{\cot 2 x}}=1$ $\lim _{x \rightarrow \frac{\pi^{-}}{2}}(1+|\cos x|)^{\frac{a|t a n x|}{b}}=e^{\frac{\lim _{T}}{x} \frac{a \sin x \mid}{b}}=e^{\frac{a}{b}}$ \(f\left(\frac{\pi}{2}\right)=\frac{\pi}{2}+3\) \(\Rightarrow \quad a=0\) D) \(\lim _{x \rightarrow 0^{0}} a+\frac{\sin [x]}{x}=a\). \(\lim _{x \rightarrow 0} b+\left[\frac{\sin x-x}{x^{3}}\right]=b-1\) \(f(0)=2\) \(\Rightarrow \mathrm{a}=2, \quad \mathrm{~b}=3\) So, \(\mathrm{T}\) is correct.

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

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

Correction $\rightarrow f(x)=\frac{1-\cos x(\cos 2 x)^{1 / 2} \cos (3 x)^{1 / 3}}{x^{2}}$ Sol using L-Hospital's Rule $\begin{aligned} & \lim _{x \rightarrow 0}(\sin x)(\cos 2 x)^{1 / 2}(\cos 3 x)^{1 / 3}+\cos x(\cos 3 x)^{1 / 3} \\ \lim _{x \rightarrow 0} f(x)=& \frac{\sin 2 x}{\sqrt{\cos 2 x}+\cos x(\cos 2 x)^{1 / 3} \frac{\sin 3 x}{(\cos 3 x)^{2 / 3}}}{2} \\\=& \lim _{x \rightarrow 0}\left[\frac{1}{2}+1+\frac{3}{2}\right]=3 \end{aligned}$
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