Nilai limit tan 2x cos 8x - tan 2x / 16x kuadrat

jawab

[tex] \lim_{x\to \ 0} \frac{tan\,2x\, cos\, 8x - tan\, x}{16 x^2} [/tex]

[tex] \lim_{x \to \ 0} \frac{tan\,2x\, (cos\, 8x - 1) }{16 x^2}[/tex]

[tex]\lim_{x \to \ 0} \frac{tan\,2x\, (-2\sin^2\, 4x) }{16 x^2}[/tex]

[tex]untuk\, x= 0\, = \frac{-2(2x)(4x)^2 }{16x^2} \,= \frac{-64 x^3}{16x^2} \,=-4x\,=4(0)=0 [/tex]