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Chapter 6: Problem 85

$\begin{aligned} \lim _{x \rightarrow \infty} f(x) &=\lim _{x \rightarrow \infty}\left(\frac{2}{x}+2+\frac{\sqrt{x^{2}+4}}{x}\right) \\ &=3 \end{aligned}$

Short Answer

Expert verified
Answer: The limit of the function as x approaches infinity is 3.

Step by step solution

01

Find the limit of the first term

We need to find the limit as x approaches infinity for the term \(\frac{2}{x}\). Since the denominator goes to infinity, the limit of the term is 0: $$\lim_{x \rightarrow \infty} \frac{2}{x} = 0$$
02

Find the limit of the second term

The second term is constant, which means its limit as x approaches infinity is the same constant: $$\lim_{x \rightarrow \infty} 2 = 2$$
03

Find the limit of the third term

We need to find the limit as x approaches infinity for the term \(\frac{\sqrt{x^{2}+4}}{x}\). To do that, we can multiply both numerator and denominator by \(\frac{1}{x}\), which gives: $$\lim_{x \rightarrow \infty} \frac{\sqrt{x^{2}+4}}{x} = \lim_{x \rightarrow \infty} \frac{\sqrt{1+\frac{4}{x^{2}}}}{1}$$ As x goes to infinity, the fraction \(\frac{4}{x^{2}}\) converges to 0: $$\lim_{x \rightarrow \infty} \frac{\sqrt{1+\frac{4}{x^{2}}}}{1} = \lim_{x \rightarrow \infty} \sqrt{1+0} = 1$$
04

Add the limits of each term

Now, we add the limits of the three terms as x approaches infinity: $$\lim_{x \rightarrow \infty} f(x) = 0 + 2 + 1 = 3$$ So, the limit of the function f(x) as x approaches infinity is 3: $$\lim_{x \rightarrow \infty} f(x) = 3$$

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Most popular questions from this chapter

\(f(x)=g(x)\) \(\sin 3 x+\cos x=\cos 3 x+\sin x\) \(\sin 3 x-\sin x=\cos 3 x-\cos x\) \(\cos 2 x \sin x=-\sin 2 x \sin x\) \(\Rightarrow \sin x(\cos 2 x+\sin 2 x)=0\) \(\Rightarrow \sin x \sin \left(\frac{\pi}{4}+2 x\right)=0\) \(\Rightarrow\) either \(\sin x=0\) or \(\sin \left(\frac{\pi}{4}+2 x\right)=0\) \(x=0, \pi, \frac{3 \pi}{8}, \frac{7 \pi}{8}\) Hence, \(\mathrm{C}\) is correct

$\mathrm{f}^{\prime}(\mathrm{x})= \begin{cases}\mathrm{x}-\\{\mathrm{x}\\}, & \mathrm{x} \geq 0 \\ -\mathrm{x}-\\{\mathrm{x}\\} & \mathrm{x}<0\end{cases}$ $=\left[\begin{array}{ll}{[\mathrm{x}],} & \mathrm{x} \geq 0 \\\ -\mathrm{x}-\\{\mathrm{x}\\}, \mathrm{x}<0\end{array}\right.$ For \(\mathrm{x} \in\left(-\frac{1}{2}, 0\right),\\{\mathrm{x}\\}=1+\mathrm{x}\) so, \(\mathrm{f}^{\prime}(\mathrm{x})=-\mathrm{x}-1-\mathrm{x}=-2 \mathrm{x}-1\) Hence, A is correct

\(f(x)=3 \cos ^{4} x+10 \cos ^{3} x+6 \cos ^{2} x-3\) \(f^{\prime}(x)=-6 \sin x\left(2 \cos ^{3} x+5 \cos ^{2} x+2 \cos x\right)\) \(=-3 \sin 2 x\left(2 \cos ^{2} x+5 \cos x+2\right)\) \(=-3 \sin 2 x(\cos x+2)(2 \cos x+1)\) Hence, \(\mathrm{C}\) is correct

$\mathrm{f}\left(\mathrm{x}_{0}\right)=\mathrm{g}\left(\mathrm{x}_{0}\right) \& \mathrm{f}^{\prime}(\mathrm{x})>\mathrm{g}^{\prime}(\mathrm{x})$ As rate of change of \(\mathrm{f}(\mathrm{x})\) is larger than \(\mathrm{g}(\mathrm{x})\) \(f(x)>g(x) \quad \forall x>x_{0}\) Hence, D is correct

\(f(x)=x \sqrt{4 a x-x^{2}}\) \(f^{\prime}(x)=\sqrt{4 a x-x^{2}}+\frac{x(2 a-x)}{\sqrt{4 a x-x^{2}}}\) \(=\frac{6 a x-2 x^{2}}{\sqrt{4 a x-x^{2}}}=\frac{2 x(3 a-x)}{\sqrt{x(4 a-x)}}\)
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