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Question Number 82800 by jagoll last updated on 24/Feb/20

$$\underset{x\to 0}{\lim }\frac{1-\sqrt{1+x^2}\cos 2x}{x^2}$$

Commented by mathmax by abdo last updated on 24/Feb/20

$$letf\left(x\right)=\frac{1-\sqrt{1+x^2}cos\left(2x\right)}{x^3}wehavecos\left(2x\right)\sim 1-2x^2$$$${(1+u)}^{\alpha }\sim 1+\alpha u+\frac{\alpha (\alpha -1)}{2}{u}^2(u\sim o)\Rightarrow$$$$\sqrt{1+x^2}\sim 1+\frac{1}{2}{x}^2+\frac{\frac{1}{2}(-\frac{1}{2})}{2}{x}^4=1+\frac{x^2}{2}-\frac{1}{8}{x}^4$$$$cosu=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{u}^{2n}}{\left(2n\right)!}\Rightarrow cosu\sim 1-\frac{u^2}{2}+\frac{u^4}{4!}\Rightarrow$$$$cos\left(2x\right)\sim 1-2x^2+\frac{{\left(2x\right)}^4}{4!}=1-2x^2+\frac{16x^4}{4.6}=1-2x^2+\frac{2}{3}{x}^4\Rightarrow$$$$\sqrt{1+x^2}cos\left(2x\right)\sim (1+\frac{x^2}{2}-\frac{x^4}{8})(1-2x^2+\frac{2}{3}{x}^4)$$$$=1-2x^2+\frac{2}{3}{x}^4+\frac{x^2}{2}-x^4+\frac{1}{3}{x}^6-\frac{x^4}{8}+\frac{1}{4}{x}^6-\frac{x^8}{12}$$$$=1+(-2+\frac{1}{2})x^2+(\frac{2}{3}-1-\frac{1}{8})x^4+(\frac{1}{3}+\frac{1}{4})x^6-\frac{x^8}{12}$$$$=1-\frac{3}{2}{x}^2+\frac{5}{24}{x}^4+\frac{7}{12}{x}^6-\frac{x^8}{12}\Rightarrow$$$$f\left(x\right)\sim \frac{\frac{3}{2}{x}^2+\frac{5}{24}{x}^4+\frac{7}{12}{x}^6-\frac{x^8}{12}}{x^3}=\frac{3}{2x}+\frac{5}{24}x+\frac{7}{12}{x}^3-\frac{1}{12}{x}^5\Rightarrow$$$${lim}_{x\to 0}f\left(x\right)=\mathrm{\infty }$$

Commented by jagoll last updated on 24/Feb/20

$$sorrymyquestionistypo.$$$$iwillcorrect$$

Answered by TANMAY PANACEA last updated on 24/Feb/20

$$\underset{x\to 0}{\lim }\frac{1-{cos}^{2}2x(1+x^2)}{x^3}\times \frac{1}{1+\sqrt{1+x^2}\times cos2x}$$$$\underset{x\to 0}{\lim }\frac{{sin}^{2}2x-x^2{cos}^{2}2x}{x^3}\times \frac{1}{1+1}$$$$\frac{1}{2}\underset{x\to 0}{\lim }\frac{(sin2x+xcos2x)}{1}\times \frac{sin2x-xcos2x}{x^3}$$$$=\frac{1}{2}\times \underset{x\to 0}{\lim }\frac{(2x-\frac{8x^3}{3!}+\frac{32x^5}{5!}..)-x(1-\frac{4x^2}{2!}+\frac{16x^4}{4!}..)}{x^3}$$$$innuminatorxcannotbeeliminated$$$$$$

Commented by jagoll last updated on 24/Feb/20

$$sirtheoriginallquestion$$$$\underset{x\to 0}{\lim }\frac{1-\sqrt{1+x^2}\cos 2x}{x^2}$$

Commented by TANMAY PANACEA last updated on 24/Feb/20

$$ok..$$

Answered by john santu last updated on 24/Feb/20

$$\underset{x\to 0}{\lim }\frac{1-(1-\frac{4x^2}{2!}+\frac{16x^4}{4!}-o\left({\left(2x\right)}^4\right))(1+\frac{1}{2}{x}^2+o\left(x^2\right))}{x^2}$$$$\underset{x\to 0}{\lim }\frac{1-(1+\frac{1}{2}{x}^2+o\left(x^2\right)-2x^2-x^4+...)}{x^2}$$$$\underset{x\to 0}{\lim }\frac{(\frac{3}{2}{x}^2-o\left(x^2\right)+x^4+...)}{x^2}=\frac{3}{2}$$

Answered by john santu last updated on 24/Feb/20

Commented by jagoll last updated on 24/Feb/20

$$yess..thankyoumister$$

Answered by Henri Boucatchou last updated on 24/Feb/20

$$\frac{1-\sqrt{1+x^2}cos2x}{x^2}=\frac{1-\sqrt{1+x^2}+2\sqrt{1+x^2}{sin}^{2}x}{x^2}$$$$=\frac{-x^2}{x^2(1+\sqrt{1+x^2})}+2\sqrt{1+x^2}{\left(\frac{sinx}{x}\right)}^2$$$$=-\frac{1}{1+\sqrt{1+x^2}}+2\sqrt{1+x^2}{\left(\frac{sinx}{x}\right)}^2\to -\frac{1}{2}+2=1,5(\mathit{x}\to 0)$$

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