Question-106956

Question Number 106956 by bemath last updated on 08/Aug/20

Answered by 1549442205PVT last updated on 08/Aug/20

((cos2x(1+sin^2 x)+cos^2 x−2)/x^2 ) =((cos2x(2−cos^2 x)+cos^2 x−2)/x^2 ) =((2(cos2x−1)+cos^2 x(cos2x+1))/x^2 ) =((−2.2sin^2 x+cos^2 x.2cos^2 x)/x^2 ) =((−4sin^2 x+2cos^4 x)/x^2 ) Hence,lim_(x→0) ((cos2x(1+sin^2 x)+cos^2 x−2)/x^2 ) =lim_(x→0) ((−4sin^2 x+2cos^4 x)/x^2 )=+∞

$$\frac{\mathrm{cos2x}\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)+\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{cos2x}\left(\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}\right)+\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}\left(\mathrm{cos2x}−\mathrm{1}\right)+\mathrm{cos}^{\mathrm{2}} \mathrm{x}\left(\mathrm{cos2x}+\mathrm{1}\right)}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{2}.\mathrm{2sin}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}.\mathrm{2cos}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{4sin}^{\mathrm{2}} \mathrm{x}+\mathrm{2cos}^{\mathrm{4}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{Hence},\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos2x}\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)+\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{4sin}^{\mathrm{2}} \mathrm{x}+\mathrm{2cos}^{\mathrm{4}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} }=+\infty \\ $$

Commented by bemath last updated on 10/Aug/20

$${wrong} \\ $$

Answered by bemath last updated on 08/Aug/20

Answered by Dwaipayan Shikari last updated on 08/Aug/20

lim_(x→0) ((cos2x(1+sin^2 x)−(1+sin^2 x))/x^2 ) lim_(x→0) −(((1+sin^2 x)(1−cos2x))/x^2 ) lim_(x→0) −2(((1+sin^2 x)sin^2 x)/x^2 )=−2(1+x^2 )(x^2 /x^2 )=−2 (sinx→x)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{cos}\mathrm{2}{x}\left(\mathrm{1}+{sin}^{\mathrm{2}} {x}\right)−\left(\mathrm{1}+{sin}^{\mathrm{2}} {x}\right)}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\frac{\left(\mathrm{1}+{sin}^{\mathrm{2}} {x}\right)\left(\mathrm{1}−{cos}\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}−\mathrm{2}\frac{\left(\mathrm{1}+{sin}^{\mathrm{2}} {x}\right){sin}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }=−\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }=−\mathrm{2}\:\:\:\left({sinx}\rightarrow{x}\right) \\ $$

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