Lim tanx-sinx / x(1-cos^2 x) x--> 0 lim 1- (akar cosx) / x^2 x--> 0 lim tan x - sin x / x^3 x--> 0 .... Question from @raynaldo13

izharulhaq78 $\displaystyle \lim_{x\to0}\frac{\tan x-\sin x}{x(1-\cos^2x)}\\\lim_{x\to0}\frac{\frac{\sin x}{\cos x}-\frac{\cos x\sin x}{\cos x}}{x(1-\cos^2x)}\\\lim_{x\to0}\frac{\sin x-\cos x\sin x}{x\cos x(1-\cos x)(1+\cos x)}\\\lim_{x\to0}\frac{\sin x(1-\cos x)}{x\cos x(1-\cos x)(1+\cos x)}\\\lim_{x\to0}\frac{x}{x\cos x(1+\cos x)}\\\lim_{x\to0}\frac{1}{\cos x(1+\cos x)}\\\\\frac{1}{\cos 0(1+\cos 0)}\\\frac{1}{1(1+1)}\\\boxed{\boxed{\frac{1}{2}}}$

$\displaystyle \lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}\\\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}\times\frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}}\\\lim_{x\to0}\frac{1^2-(\sqrt{\cos x})^2}{x^2(1+\sqrt{\cos x})}\\\lim_{x\to0}\frac{1-\cos x}{x^2(1+\sqrt{\cos x})}\\\lim_{x\to0}\frac{\frac{\sin^2x}{1+\cos x}}{x^2(1+\sqrt{\cos x})}\\\lim_{x\to0}\frac{\sin^2x}{x^2(1+\sqrt{\cos x})(1+\cos x)}\\\lim_{x\to0}\frac{\sin x\times\sin x}{x^2(1+\sqrt{\cos x})(1+\cos x)}\\\lim_{x\to0}\frac{1}{(1+\sqrt{\cos x})(1+\cos x)}$

$\displaystyle \frac{1}{(1+\sqrt{\cos 0})(1+\cos 0)}\\\frac{1}{(1+1)(1+1)}\\\frac{1}{(2)(2)}\\\boxed{\boxed{\frac{1}{4}}}$

$\displaystyle \lim_{x\to0}\frac{\tan x-\sin x}{x^3}\\\lim_{x\to0}\frac{\frac{\sin x}{\cos x}-\frac{\cos x\sin x}{\cos x}}{x^3}\\\lim_{x\to0}\frac{\sin x-\cos x\sin x}{x^3\cos x}\\\lim_{x\to0}\frac{\sin x(1-\cos x)}{x^3\cos x}\\\lim_{x\to0}\frac{1-\cos x}{x^2\cos x}\\\lim_{x\to0}\frac{\frac{\sin^2x}{1+\cos x}}{x^2\cos x}\\\lim_{x\to0}\frac{\sin x\times\sin x}{x^2\cos x(1+\cos x)}\\\lim_{x\to0}\frac{1}{\cos x(1+\cos x)}\\\\\frac{1}{\cos 0(1+\cos0)}\\\frac{1}{1(1+1)}\\\boxed{\boxed{\frac{1}{2}}}$

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raynaldo13 yg punya kau benar

Sa1j Nomor 1
$$\begin {align}\lim_{x \to 0} \frac{\tan x-\sin x}{x(1-\cos^2x)}&=\lim_{x \to 0} \frac{\sin x(\frac{1}{\cos x}-1)}{x\sin^2x}\\&=\lim_{x \to 0} \frac{1-\cos x}{x\sin x\cos x}\\&=\lim_{x \to 0} \frac{2\sin^2\frac{1}{2}x}{x\sin x\cos x}\\&=2.\frac{1}{2}.\frac{1}{2}.\frac{1}{\cos0}\\&=\boxed{\boxed{\bold{\frac{1}{2}}}}
\end$

Nomor 2
$$\begin {align} \lim_{x \to0} \frac{1-\sqrt{\cos x}}{x^2}&=\lim_{x \to0} \frac{1-\sqrt{\cos x}}{x^2}\times \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}}\\&=\lim_{x \to0} \frac{1-\cos x}{x^2(1+\sqrt{\cos x})}\\&=\lim_{x \to0} \frac{2\sin^2\frac{1}{2}x}{x^2(1+\sqrt{\cos x})}\\&=2.\frac{1}{2}.\frac{1}{2}..\frac{1}{2}\\&=\boxed{\boxed{\bold{\frac{1}{4}}}}
\end$

Nomor 3
$$\begin {align}\lim_{x \to 0} \frac{\tan x-\sin x}{x^3}&=\lim_{x \to 0} \frac{\sin x(\frac{1}{\cos x}-1)}{x^3}\\&=\lim_{x \to 0} \frac{(1-\cos x)}{x^2\ \cos x}\\&=\lim_{x \to 0} \frac{2\sin^2\frac{1}{2}x}{x^2\ \cos x}\\&=2.\frac{1}{2}.\frac{1}{2}.\frac{1}{1}\\&=\boxed{\boxed{\bold{\frac{1}{2}}}}
\end$

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Sa1j udah deh setiap ke depan ada pertanyaan dari dia diignore aja

raynaldo13 Nomor 1 [tex]$\begin {align}\lim_{x \to 0} \frac{\tan x-\sin x}{x(1-\cos^2x)}&=\lim_{x \to 0} \frac{\sin x(\frac{1}{\cos x}-1)}{x\sin^2x}\\&=\lim_{x \to 0} \frac{1-\cos x}{x\sin x\cos x}\\&=\lim_{x \to 0} \frac{2\sin^2\frac{1}{2}x}{x\sin x\cos x}\\&=2.\frac{1}{2}.\frac{1}{2}.\frac{1}{\cos0}\\&=\boxed{\boxed{\bold{\frac{1}{2}}}} \end [/tex]

raynaldo13 jawaban apo yg kau kayak gini

raynaldo13 Nomor 2 [tex]$\begin {align} \lim_{x \to0} \frac{1-\sqrt{\cos x}}{x^2}&=\lim_{x \to0} \frac{1-\sqrt{\cos x}}{x^2}\times \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}}\\&=\lim_{x \to0} \frac{1-\cos x}{x^2(1+\sqrt{\cos x})}\\&=\lim_{x \to0} \frac{2\sin^2\frac{1}{2}x}{x^2(1+\sqrt{\cos x})}\\&=2.\frac{1}{2}.\frac{1}{2}..\frac{1}{2}\\&=\boxed{\boxed{\bold{\frac{1}{4}}}} \end[/tex]

Sa1j wah saya paham berarti kamu melihatnya dari android.. udah deh gk usah di perpanjang... coba kamu lihat aja pake komputer

raynaldo13 Nomor 3 [tex]$\begin {align}\lim_{x \to 0} \frac{\tan x-\sin x}{x^3}&=\lim_{x \to 0} \frac{\sin x(\frac{1}{\cos x}-1)}{x^3}\\&=\lim_{x \to 0} \frac{(1-\cos x)}{x^2\ \cos x}\\&=\lim_{x \to 0} \frac{2\sin^2\frac{1}{2}x}{x^2\ \cos x}\\&=2.\frac{1}{2}.\frac{1}{2}.\frac{1}{1}\\&=\boxed{\boxed{\bold{\frac{1}{2}}}} \end[/tex]

Sa1j kalau kamu pake komputer coba direfresh aja supaya gk ada kesalahpahaman diantara kit

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Agar didapatkan larutan dengan pH= 5, maka kedalam 500 ml air harus dilarutkan (NH4)2SO4 Kb= 10^-5, Mr= 132 sebanyak........gram. TOLONG DIJAWAB DENGAN PENJELSAN YANG DIMENGERTI BAGI YANG MENGERTI DAN MINTA BANTUAN BAGI MODERATOR

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