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Question Number 105614 by bemath last updated on 30/Jul/20

lim_(x→0) ((tan (cos 2x−1))/(2x^2 )) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left(\mathrm{cos}\:\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} }\:? \\ $$

Commented by PRITHWISH SEN 2 last updated on 30/Jul/20

let cos2x−1=t⇒x→0 then t→0 lim_(t→0) ((tan t)/t) lim_(x→0) ((cos 2x−1)/(2x^2 )) = lim_(x→0) −(((sin x)/x))^2 = −1

$$\mathrm{let}\:\mathrm{cos2x}−\mathrm{1}=\mathrm{t}\Rightarrow\mathrm{x}\rightarrow\mathrm{0}\:\mathrm{then}\:\mathrm{t}\rightarrow\mathrm{0} \\ $$$$\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{t}}{\mathrm{t}}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{cos}\:\mathrm{2x}−\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:−\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{x}}\right)^{\mathrm{2}} =\:−\mathrm{1} \\ $$

Answered by Dwaipayan Shikari last updated on 30/Jul/20

lim_(x→0) −((tan(2sin^2 x))/(2x^2 ))=lim_(x→0) −((tan(2x^2 ))/(2x^2 ))=−((2x^2 )/(2x^2 ))=−1 2sin^2 x →2x^2 tan(2x^2 )→2x^2

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

Answered by bobhans last updated on 30/Jul/20

lim_(x→0) ((tan (1−(4/2)x^2 −1))/(2x^2 ))= lim_(x→0) ((tan (−2x^2 ))/(2x^2 )) = −1 ★

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left(\mathrm{1}−\frac{\mathrm{4}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{x}^{\mathrm{2}} }=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\:\left(−\mathrm{2}{x}^{\mathrm{2}} \right)}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$=\:−\mathrm{1}\:\bigstar \\ $$

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