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Chapter 2: Problem 60

\(f(x)=\lim _{x \rightarrow \infty} x \tan ^{-1} n x\) $\left\\{\begin{array}{cl}\frac{\pi}{2} x, & x>0 \\ 0, & x=0 \\\ \frac{-\pi}{2} x, & x<0\end{array}\right.$

Short Answer

Expert verified
Answer: The limit of the function \(f(x) = x\tan^{-1}(nx)\) as x approaches infinity depends on the value of x: - If \(x > 0\), the limit is \(\infty\). - If \(x = 0\), the limit is \(0\). - If \(x < 0\), the limit is \(-\infty\).

Step by step solution

01

Determine the limit of the function given its definition

First, we need to understand what the given function definition means. It says that \(f(x)\) is equal to the limit as x approaches infinity of \(x\tan^{-1}(nx)\). So, our objective is to find the value of this limit as x approaches infinity.
02

Split the function based on the value of x

According to the function definition, we have three different expressions for \(f(x)\), depending on the value of x: - If \(x > 0\), then \(f(x) = \frac{\pi}{2}x\). - If \(x = 0\), then \(f(x) = 0\). - If \(x < 0\), then \(f(x) = \frac{-\pi}{2}x\). Now, we will compute the limit of each of the expressions as x approaches infinity.
03

Calculate the limit of each expression

First, let's evaluate the limit for x > 0: \(\lim_{x\rightarrow\infty}\frac{\pi}{2}x = \infty\) Next, let's evaluate the limit for x = 0: \(\lim_{x\rightarrow\infty}0 = 0\) Finally, let's evaluate the limit for x < 0: \(\lim_{x\rightarrow\infty}\frac{-\pi}{2}x = -\infty\)
04

Combine the results

Now that we have evaluated the limits for all three expressions, we can conclude the final function as follows: \[ f(x) = \lim_{x\rightarrow\infty}x\tan^{-1}(nx) = \left\{ \begin{array}{cl} \infty, & x > 0 \\ 0, & x = 0 \\ -\infty, & x < 0 \end{array} \right. \]

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Most popular questions from this chapter

$\lim _{x \rightarrow 0} \frac{\sin ^{4}\left(\frac{1}{x}\right)-\sin ^{2}\left(\frac{1}{x}\right)+1}{\cos ^{4}\left(\frac{1}{x}\right)-\cos ^{2}\left(\frac{1}{x}\right)+1}$ $\lim _{x \rightarrow 0} \frac{-\sin ^{2}\left(\frac{1}{x}\right) \cos ^{2}\left(\frac{1}{x}\right)+1}{-\cos ^{2}\left(\frac{1}{x}\right) \sin ^{2}\left(\frac{1}{x}\right)+1}=1$

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

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.

$\prod_{n=1}^{n} \cos \left(\frac{x}{2^{n}}\right)=\frac{\sin x}{2^{n} \sin \left(\frac{x}{2^{n}}\right)}$ 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, $\lim _{n \rightarrow \infty} \sum_{n=1}^{n} \frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{1}{x}-\cot x$ $\lim _{x \rightarrow \frac{\pi}{2}} \lim _{n \rightarrow \infty} \sum_{n=1}^{n} \frac{1}{2^{n}} \tan \left(\frac{x}{2^{n}}\right)=\frac{2}{\pi}$

$\lim _{x \rightarrow a}(\cos (x-a))^{\frac{6}{(x-a) 2 \sin (x-a) \cos (x-a)}}$ $\Rightarrow e^{-\lim 2 \sin ^{2} \frac{(x-a)}{2} \times \frac{6}{(x-a) \sin (x-a) \cos (x-a)}}$ \(\Rightarrow e^{-3 / 2}\) Now, $h(x)=\lim _{n \rightarrow \infty} \frac{\sin \frac{x}{2^{n}}}{\frac{x}{2^{n}}} \times x=x$ \(h(a)=a\) \(\Rightarrow \mathrm{k}=\frac{3}{2}\) Hence, \(\mathrm{C}\) is correct.
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