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Chapter 2: Problem 95

$\lim _{x \rightarrow 0^{\prime}} \cos \left(x \cos \frac{1}{x}\right)=1 \quad f(0)$ is not defined. \(\lim _{x \rightarrow 0^{-}} \cos \left(x \cos \frac{1}{x}\right)=1\)

Short Answer

Expert verified
Answer: The one-sided limits of the function are both equal to 1. That is, \(\lim_{x \rightarrow 0^+} f(x) = 1\) and \(\lim_{x \rightarrow 0^-} f(x) = 1\).

Step by step solution

01

Identify the function and the one-sided limits

We have to find the one-sided limits of the function \(f(x) = \cos \left(x \cos \frac{1}{x}\right)\) as \(x\) approaches \(0\). We are given that \(\lim_{x \rightarrow 0^+} f(x) = 1\) and have to find the limit as \(x \rightarrow 0^-\).
02

Analyze the function as \(x\) approaches \(0^-\)

As \(x\) approaches \(0\) from the left, the term \(x\cos \frac{1}{x}\) approaches \(0\) as well, regardless of the value of the cosine function. This is because the \(x\) term dominates the behavior of the entire expression.
03

Use the properties of the cosine function

As the argument of the cosine function approaches \(0\), the cosine of the argument approaches \(1\). That is, \(\lim_{x \rightarrow 0} \cos x = 1\).
04

Apply the limit properties to find the limit from the left

Now we can apply the limit properties to the function \(f(x)\) as \(x\) approaches \(0\) from the left: $$ \lim_{x \rightarrow 0^-} \cos \left(x \cos \frac{1}{x}\right) = \cos \left(\lim_{x \rightarrow 0^-} x \cos \frac{1}{x}\right). $$ Since we know that the limit of the argument is \(0\), we get, $$ \cos \left(\lim_{x \rightarrow 0^-} x \cos \frac{1}{x}\right) = \cos(0) = 1. $$
05

Final answer

We have now shown that the limit of the function \(f(x) = \cos \left(x \cos \frac{1}{x}\right)\) as \(x\) approaches \(0\) from the left is also \(1\): $$ \lim_{x \rightarrow 0^-} \cos \left(x \cos \frac{1}{x}\right) = 1. $$

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Most popular questions from this chapter

$f\left(\frac{1}{4^{n}}\right)=\left(\sin e^{n}\right) \mathrm{e}^{-n^{2}}+\frac{n^{2}}{n^{2}+1}$ $f(0)=\lim _{n \rightarrow \infty}\left(\left(\sin e^{n}\right) e^{-n^{i}}+\frac{n^{2}}{n^{2}+1}\right)$ \(=1\)

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

$f(x)= \begin{cases}\frac{[x]+\sqrt{x-[x]}}{\sin x}, & x \geq 0 \\ \sin x, & x<0\end{cases}$ If \(x\) is te integer \(=k\) \(f(x)=k\) $\lim _{x \rightarrow k^{+}} f(x)=k, \quad \lim _{x \rightarrow k} f(x)=(k-1)+1=k$ If \(x\) is \(-\) ve integer \(=K\) \(f(x)=\sin k\) \(\mathrm{LHL}=\mathrm{RHL}=\sin \mathrm{K}\) If \(x=0\) \(f(0)=0=R H L, L H L=\sin 0=0\) If \(x\) is any \(+\) real \(n o,=K\) \(\mathrm{f}(\mathrm{x})=[\mathrm{k}]+\sqrt{\\{\mathrm{k}\\}}\) $\mathrm{RHL}=[\mathrm{k}]+\sqrt{\\{\mathrm{k}\\}}, \quad \mathrm{LHL}=[\mathrm{K}]+\sqrt{\\{\mathrm{k}\\}}$ If \(x\) is any \(-\) Real \(n_{0}=K\) $\mathrm{f}(\mathrm{x})=\sin \mathrm{K}=\mathrm{RHL}=\mathrm{LHL}$

$\lim _{x \rightarrow 1^{\prime}} g(x)=\lim _{x \rightarrow 1^{-} n \rightarrow \infty} \lim _{x \rightarrow \infty} \frac{x+x^{2 n} \sin x}{1+x^{4 n}}$ $\quad=\lim _{x \rightarrow 1^{-}} \lim _{n \rightarrow \infty} \frac{x^{-4 n+1}+x^{-2 n} \sin x}{x^{-4 n}+1}$ \(=0\) $\lim _{x \rightarrow 1} g(x)=\lim _{x \rightarrow 1} \lim _{n \rightarrow \infty} \frac{x+x^{2 n} \sin x}{1+x^{4 n}}=1$

$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}+1}{\mathrm{x}}, \quad \mathrm{x} \neq 0$ $f \circ f(x)=\frac{\frac{x+1}{x}+1}{\frac{x+1}{x}}=\frac{2 x+1}{x+1}, x \neq-1$ \(f \circ f \circ f(x)=\frac{\frac{2 x+1}{x+1}+1}{\frac{2 x+1}{x+1}}\) \(=\frac{3 x+2}{2 x+1}, \quad x \neq \frac{-1}{2}\) So, there are three points of discontinuity
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