Question Number 10398 by amir last updated on 07/Feb/17

Answered by arge last updated on 07/Feb/17

por l′hopital,    y= (1/x^2 ) − (1/(tan^2 x))    y=x^(−2)   − ((cos^2 x)/(sen^2 x))    y′= −2x^(−3) −((sen^2 x(−2senxcosx)−cos^2 x(2senxcosx))/(sen^4 x))    y′= −2x^(−3)  − (A/((1−cos^2 x)^2 ))    y′=0/∞=0

Answered by mrW1 last updated on 09/Feb/17

lim_(x→0) ((1/x^2 )−(1/(tan^2  x)))  =lim_(x→0) [((1/x)+((cos x)/(sin x)))((1/x)−((cos x)/(sin x)))]  =lim_(x→0) [((sin x+xcos x)/(xsin x))×((sin x−xcos x)/(xsin x))]  =lim_(x→0) [((sin x+xcos x)/(xsin^2  x))×((sin x−xcos x)/x)]  =lim_(x→0) [(((((sin )/x))+cos x)/((((sin x)/x))^2 ))×((sin x−xcos x)/x^3 )]  =lim_(x→0) (((((sin )/x))+cos x)/((((sin x)/x))^2 ))×lim_(x→0) ((sin x−xcos x)/x^3 )  =((1+1)/1^2 )×lim_(x→0) ((sin x−xcos x)/x^3 )  =2×lim_(x→0) ((sin x−xcos x)/x^3 )  =2×lim_(x→0) ((cos x−cos x+xsin x)/(3x^2 ))       ← applying L′hopital′s rule: lim_(x→c)   ((f(x))/(g(x)))=lim_(x→c)  ((f′(x))/(g′(x)))  =2×lim_(x→0) ((xsin x)/(3x^2 ))        =2×lim_(x→0) ((sin x)/(3x))  =(2/3)×lim_(x→0) ((sin x)/x)  =(2/3)×1  =(2/3)

Commented byamir last updated on 09/Feb/17

thank you dear mrW1.