/*! 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 74 \(f(x)=\lim _{n \rightarrow \inf... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 74

\(f(x)=\lim _{n \rightarrow \infty} \frac{x^{2 n-1}+a x^{2}+b x}{x^{2 n}+1}\) \(\lim _{x \rightarrow 1^{+}} f(x)=1\) \(\lim _{x \rightarrow 1^{-}} f(x)=a+b\) \(f(1)=\frac{a+b+1}{2}\) \(\Rightarrow a+b=1\)

Short Answer

Expert verified
We have found that under the given conditions, the function has the values \(a = 1 - b\) and \(b = 1 - a\). These can be written as one equation, \(a + b = 1\), which represents a linear relationship between the parameters \(a\) and \(b\). This means that there is not a unique solution for \(a\) and \(b\), and there can be multiple pairs of values that will satisfy the given conditions.

Step by step solution

01

We know that the limit is equal to 1, so we can write \(\lim _{x \rightarrow 1^{+}} f(x) = 1\). In order to find the correct limit expression for \(f(x)\), we must consider the value of \(x\). If \(x \rightarrow 1^+\), then \(x > 1\). Since \(x^{2n}\) will dominate the numerator and denominator, we can simplify the limit expression as follows: $$\lim _{n \rightarrow \infty} \frac{x^{2 n-1}+a x^{2}+b x}{x^{2 n}+1}=\lim _{n \rightarrow \infty} \frac{x^{2 n-1}}{x^{2 n}} =\lim _{n \rightarrow \infty} x^{-1}=1$$ #Step 2: Calculate the limit of the function as \(x \rightarrow 1^-\)#

Now we find the limit as \(x \rightarrow 1^{-}\), which means \(x < 1\). Since \(x^{2n}\) will dominate the numerator and denominator, we can simplify the limit expression as follows: $$\lim _{n \rightarrow \infty} \frac{x^{2 n-1}+a x^{2}+b x}{x^{2 n}+1}=\lim _{n \rightarrow \infty} \frac{a x^{2}+b x}{1} = a+b$$ #Step 3: Evaluate the function at \(x=1\)#
02

Now, we substitute \(x=1\) into the function definition: $$f(1)=\lim _{n \rightarrow \infty} \frac{1^{2 n-1}+a \cdot 1^{2}+b \cdot 1}{1^{2 n}+1}=\frac{a+b+1}{2}$$ #Step 4: Solve for \(a\) and \(b\)#

Now we have three equations: 1. \(\lim _{x \rightarrow 1^{+}} f(x)=1\), which implies \(1=1\) 2. \(\lim _{x \rightarrow 1^{-}} f(x)=a+b\) 3. \(f(1)=\frac{a+b+1}{2}\) We can use the last two equations to solve for \(a\) and \(b\): $$a+b=1 \Rightarrow b = 1-a$$ $$f(1)=\frac{a+b+1}{2} \Rightarrow 2f(1)=a+b+1 \Rightarrow a+b=1$$ Thus, we have the consistent equations: $$a+b=1,\quad b=1-a$$ Substituting for one of the variables gives: $$a+(1-a)=1 \Rightarrow a=1-b \Rightarrow b=1-a$$ And the unique solution is: $$a=1-b$$ $$b=1-a$$ So, the given condition has been satisfied with the values \(a=1-b\) and \(b=1-a\).

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Most popular questions from this chapter

$\lim _{x \rightarrow \frac{\pi^{-}}{2}}\left[\frac{2\left(\sin x-\sin ^{3} x\right)+\sin x-\sin ^{3} x}{2\left(\sin x-\sin ^{3} x\right)-\sin x+\sin ^{3} x}\right]$ $\lim _{n \rightarrow \frac{\pi}{2}}\left[\frac{3\left(\sin x-\sin ^{3} x\right)}{\left(\sin x-\sin ^{3} x\right)}\right]=3$ $\lim _{x \rightarrow \frac{\pi}{2}}\left[\frac{2\left(\sin x-\sin ^{3} x\right)+\sin x-\sin ^{3} x}{2\left(\sin x-\sin ^{3} x\right)-\sin x+\sin ^{3} x}\right]$ \(=3\) So, \(f(x)\) is continuous at \(x=\frac{\pi}{2}\) LHD \(=\) RHD \(=0\) (By first principle)

\(\mathrm{f}(\mathrm{x})=\) highest power of \(\left(\mathrm{u}^{\mathrm{x}^{2}}+\mathrm{u}^{2}+2 \mathrm{u}+3\right)\) $\lim _{\mathrm{x} \rightarrow \sqrt{2}} \mathrm{f}(\mathrm{x})=2 \quad \mathrm{f}(\sqrt{2})=2$ \(\lim _{x \rightarrow \sqrt{2}} \mathrm{f}(\mathrm{x})=2\)

$f(x)=\frac{\log _{\sin x} \cos ^{3} x}{\log _{\sin |3 x|} \cos \frac{x}{2}}, x \in\left(-\frac{\pi}{3}, \frac{\pi}{3}\right), \quad x \neq 0$ $=\frac{3 \log (1+\cos x-1)}{\log \left(1+\cos \frac{x}{2}-1\right)} \times \frac{\log (1+\sin |3 x|-1)}{\log (1+\sin |x|-1)}$ \(f(x)\) is discontinuous at \(x=\pm \frac{\pi}{6}\)

If \(x \geq 0\) \(x^{2}-x+1-2=0\) \(x^{2}-x-1=0\) \(x=1 \pm \sqrt{5} \Rightarrow x=\frac{1+\sqrt{5}}{2}\) If \(x<0\) \(x^{2}-x-2=0\) \((x-2)(x+1)=0\) \(\mathrm{x}=2\) or \(-1\) so, \(x=-1\) No options matches.

$f(x)=\left\\{\begin{array}{lr}2 x-1, & -2 \leq x<0 \\ x+2, & 0 \leq x \leq 2\end{array}\right.$ $g(x)= \begin{cases}{[x],} & -4 \leq x<-2 \\ x+2, & -2 \leq x \leq 4\end{cases}$ \(\lim _{x \rightarrow-2^{2}} f(g(x))=2\) \(\lim _{x \rightarrow-2} f(g(x)=\) Not defined
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