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Question Number 65356 by aliesam last updated on 28/Jul/19

Commented by mathmax by abdo last updated on 29/Jul/19

$$letA\left(x\right)=\frac{\sqrt{{cos}^{4}x-cos\left(2x\right)}}{ln(1+3x)}wehave$$$${cos}^{4}x-cos\left(2x\right)={\left(\frac{1+cos\left(2x\right)}{2}\right)}^2-cos\left(2x\right)$$$$=\frac{1}{4}(1+2cos\left(2x\right)+{cos}^2\left(2x\right))-cos\left(2x\right)$$$$=\frac{1}{4}-\frac{1}{2}cos\left(2x\right)+\frac{1}{8}(1+cos\left(4x\right))=\frac{3}{8}-\frac{1}{2}cos\left(2x\right)+\frac{1}{8}cos\left(4x\right)$$$$cos\left(2x\right)\sim 1-\frac{{\left(2x\right)}^2}{2}+\frac{{\left(2x\right)}^4}{4!}=1-2x^2+\frac{2}{3}{x}^4\Rightarrow$$$$\frac{1}{2}cos\left(2x\right)\sim \frac{1}{2}-x^2+\frac{1}{3}{x}^4$$$$cos\left(4x\right)\sim 1-\frac{{\left(4x\right)}^2}{2}+\frac{{\left(4x\right)}^4}{4!}=1-8x^2+\frac{4^3}{6}{x}^4\Rightarrow$$$$\frac{1}{8}cos\left(4x\right)\sim \frac{1}{8}-x^2+\frac{4^3}{48}{x}^4\Rightarrow$$$${cos}^{4}x-cos\left(2x\right)\sim \frac{3}{8}-\frac{1}{2}+x^2-\frac{x^4}{3}+\frac{1}{8}-x^2+\frac{4^3}{48}{x}^4$$$$=(\frac{4^3}{48}-\frac{1}{3})x^4\Rightarrow \sqrt{{cos}^{4}x-cos\left(2x\right)}\sim \sqrt{(\frac{4^3}{48}-\frac{1}{3})}{x}^{2}alsowehave$$$$ln(1+3x)\sim 3x\Rightarrow A\left(x\right)\sim \lambda x(x\in V\left(o\right))\Rightarrow {lim}_{x\to 0}A\left(x\right)=0$$$$$$

Answered by Tanmay chaudhury last updated on 30/Jul/19

$$ormethod$$$$N_r=\sqrt{{cos}^{4}x-(2{cos}^{2}x-1)}$$$$=\sqrt{{({cos}^{2}x-1)}^2}$$$$=\sqrt{{(1-{cos}^{2}x)}^2}$$$$={sin}^{2}x$$$$\underset{x\to 0}{\lim }\frac{{sin}^{2}x}{ln(1+3x)}$$$$\underset{x\to 0}{\lim }\frac{{\left(\frac{sinx}{x}\right)}^2\times x^2}{\frac{ln(1+3x)}{3x}\times 3x}$$$$=\underset{x\to 0}{\lim }\frac{1^2\times {\mathit{x}}^2}{1\times 3\mathit{x}}$$$$=\underset{x\to 0}{\lim }\frac{\mathit{x}}{3}$$$$=0$$$$\mathit{e}\mathit{d}\mathit{i}\mathit{t}\mathit{e}\mathit{d}$$$$=\frac{1\times 0}{1\times 3}=0$$$$$$$$$$$$$$

Commented by Tanmay chaudhury last updated on 29/Jul/19

$$yessiryouarecorrect$$

Commented by kaivan.ahmadi last updated on 29/Jul/19

$$why1\times 3?itmustbe1\times 0$$

Commented by mathmax by abdo last updated on 29/Jul/19

$${lim}_{x\to 0}\frac{{sin}^{2}x}{ln(1+3x)}={lim}_{x\to 0}\frac{{\left(\frac{sinx}{x}\right)}^2\times x^2}{\frac{ln(1+3x)}{3x}\times 3x}$$$$={lim}_{x\to 0}\frac{x^2}{3x}={lim}_{x\to 0}\frac{x}{3}=0because{lim}_{x\to 0}\frac{sinx}{x}=1$$$$and{lim}_{x\to 0}\frac{ln(1+3x)}{3x}=1$$

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