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Question Number 203288 by hardmath last updated on 14/Jan/24

$$\mathrm{Find}:\underset{\mathbf{x}\to 0}{\lim }\frac{1-\cos 4\mathrm{x}}{{\mathrm{x}}^2}=?$$

Answered by MM42 last updated on 14/Jan/24

$$={lim}_{x\to 0}\frac{2{sin}^{2}2x}{x^2}$$$$={lim}_{x\to 0}8{\left(\frac{sin2x}{2x}\right)}^2=8✓$$

Commented by hardmath last updated on 14/Jan/24

$$\mathrm{thankyou}\mathrm{dearprofessor}$$

Answered by Calculusboy last updated on 15/Jan/24

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:\mathit{N}\mathit{B}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{1-\mathit{c}\mathit{o}\mathit{s}\mathit{a}\mathit{x}}{{\mathit{x}}^2}=\frac{{\mathit{a}}^2}{2}$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{n};\mathit{l}\mathit{i}\underset{\mathit{x}\to 0}{\mathit{m}}\frac{1-\mathit{c}\mathit{o}\mathit{s}4\mathit{x}}{{\mathit{x}}^2}=\frac{4^2}{2}=\frac{16}{2}=8$$$$\mathit{M}\mathit{f}$$

Answered by dimentri last updated on 15/Jan/24

$$=\underset{x\to 0}{\lim }\frac{1-\cos {}^{2}4\mathrm{x}}{{\mathrm{x}}^2(1+\cos 4\mathrm{x})}$$$$=\underset{x\to 0}{\lim }{\left(\frac{\sin 4\mathrm{x}}{\mathrm{x}}\right)}^2.\underset{x\to 0}{\lim }\frac{1}{1+\cos 4\mathrm{x}}$$$$=16\times \frac{1}{2}=8$$

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