Tolong Dibantu Dong Kak Perlu Banget Soalnya
lim x->1 tan (x-1) sin(√x-1)/ x² - 2x+1 =
lim x->1 tan (x-1) sin(√x-1)/ x² - 2x+1 =
Limit x mendekati 0
triGonometri
Penjelasan dengan langkah-langkah:
[tex]\sf \lim_{x\to 1}\ \dfrac{tan(x- 1).\sin (\sqrt x - 1) }{x^2-2x+ 1}[/tex]
[tex]\sf \lim_{x\to 1}\ \dfrac{tan(x- 1).\sin (\sqrt x - 1) }{(x -1)(x-1)}[/tex]
[tex]\sf \lim_{x\to 1}\ \dfrac{tan(x- 1).\sin (\sqrt x - 1) }{(x -1)(\sqrt x-1)(\sqrt x + 1)}[/tex]
[tex]\sf \lim_{x\to 1}\ \dfrac{1}{\sqrt x + 1} = \dfrac{1}{2}[/tex]
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