lim_(x→1) (tan(x−1)sin(1−√(x)))/(x²−2x+1)=⋯ A. 1 B. 1/2 C. 0 D. −1/2 E. −1

Jawaban D Ingat bahwa! lim_(x→0) (sin ax)/(ax) = 1 lim_(x→0) (tan ax)/(ax) = 1 lim_(x→a) f(x)·g(x) = lim_(x→a) f(x)·lim_(x→a)g(x) misalkan y = x -1, maka x = y +1 Saat x mendekati 1, maka y mendekati 0 Nilai dari lim_(x→1) (tan(x−1)sin(1−√x))/(x²−2x+1) sebagai berikut lim_(x→1) (tan(x−1)sin(1−√x))/(x²−2x+1) =lim_(y→0) (tan(y +1- 1 )sin(1−√(y +1)))/((y +1)²−2(y +1)+1)) =lim_(y→0) (tan y sin(1-√(y+1)))/(y²+2y+1-2y-2+1) =lim_(y→0) (tan y sin(1-√(y+1)))/(y²) =lim_(y→0) (tan y ·sin(1−√(y+1))/(y·y) =lim_(y→0) (tan y/(y)· lim_(y→0)(sin(1-√(y+1)))/(y+1)-1) =lim_(y→0) (tan y)/(y) · lim_(y→0)(-sin(√(y+1)-1))/((√(y+1)-1)(√(y+1)+1)) = lim_(y→0) (tan y)/(y) · lim_(y→0)(-sin(√(y+1)-1)/((√(y+1)-1) · lim_(y→0) 1/(√(y+1)+1) =1·(-1)·(1/(√1 +1)) = -1· (1/(1+1) = -1/2 Jadi, jawaban yang tepat adalah D