Jika lim x->0 sinx/x =1 maka nilai dari lim x->0 (2/x^2 -

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Pembahasan

Untuk menyelesaikan limit $\lim_{x \to 0} \left(\frac{2}{x^2} - \frac{\sin 2x}{x^2 \tan x}\right)$, kita dapat menggunakan identitas trigonometri dan aturan L'Hopital atau ekspansi deret Taylor. 

Menggunakan ekspansi deret Taylor di sekitar x=0:
$\sin 2x \approx 2x - \frac{(2x)^3}{3!} = 2x - \frac{8x^3}{6} = 2x - \frac{4x^3}{3}$
$\tan x \approx x + \frac{x^3}{3}$

Maka, \frac{\sin 2x}{x^2 \tan x} \approx \frac{2x - \frac{4x^3}{3}}{x^2 \left(x + \frac{x^3}{3}\right)} = \frac{2x(1 - \frac{2x^2}{3})}{x^3(1 + \frac{x^2}{3})} = \frac{2}{x^2} \frac{(1 - \frac{2x^2}{3})}{(1 + \frac{x^2}{3})}

Sehingga, \frac{2}{x^2} - \frac{\sin 2x}{x^2 \tan x} \approx \frac{2}{x^2} - \frac{2}{x^2} \frac{(1 - \frac{2x^2}{3})}{(1 + \frac{x^2}{3})} = \frac{2}{x^2} \left(1 - \frac{(1 - \frac{2x^2}{3})}{(1 + \frac{x^2}{3})}\right)

 = \frac{2}{x^2} \left(\frac{(1 + \frac{x^2}{3}) - (1 - \frac{2x^2}{3})}{(1 + \frac{x^2}{3})}\right) = \frac{2}{x^2} \left(\frac{\frac{x^2}{3} + \frac{2x^2}{3}}{(1 + \frac{x^2}{3})}\right) = \frac{2}{x^2} \left(\frac{x^2}{(1 + \frac{x^2}{3})}\right) = \frac{2}{(1 + \frac{x^2}{3})}

Ketika x -> 0, \frac{2}{(1 + \frac{x^2}{3})} -> \frac{2}{(1 + 0)} = 2.

Cara lain menggunakan limit yang diberikan $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Kita bisa manipulasi ekspresi:
$\frac{\sin 2x}{x^2 \tan x} = \frac{\sin 2x}{2x} \cdot \frac{2x}{x^2} \cdot \frac{x}{\tan x} \cdot \frac{1}{x} = \frac{\sin 2x}{2x} \cdot 2 \cdot \frac{x}{\tan x} \cdot \frac{1}{x}$
Kita tahu $\lim_{x \to 0} \frac{\sin 2x}{2x} = 1$ dan $\lim_{x \to 0} \frac{x}{\tan x} = 1$.

Maka, \frac{2}{x^2} - \frac{\sin 2x}{x^2 \tan x} = \frac{2}{x^2} - \frac{\sin 2x}{x^2 \tan x} \cdot \frac{2}{2} \cdot \frac{\tan x}{x} \cdot x = \frac{2}{x^2} - \frac{\sin 2x}{2x} \cdot \frac{2}{x \tan x} \cdot x = \frac{2}{x^2} - \frac{\sin 2x}{2x} \cdot \frac{2 \tan x}{x^2 \tan x} = \frac{2}{x^2} - \frac{\sin 2x}{2x} \frac{2 \tan x}{x^2 \tan x}

Mari kita ubah ekspresinya menjadi:
$\frac{2}{x^2} - \frac{\sin 2x}{x^2 \tan x} = \frac{2 \tan x - \sin 2x}{x^2 \tan x}$
Kita gunakan $\sin 2x = 2 \sin x \cos x$ dan $\tan x = \frac{\sin x}{\cos x}$:
$\frac{2 \frac{\sin x}{\cos x} - 2 \sin x \cos x}{x^2 \frac{\sin x}{\cos x}} = \frac{\frac{2 \sin x - 2 \sin x \cos^2 x}{\cos x}}{\frac{x^2 \sin x}{\cos x}} = \frac{2 \sin x (1 - \cos^2 x)}{x^2 \sin x} = \frac{2 \sin x \sin^2 x}{x^2 \sin x} = \frac{2 \sin^2 x}{x^2}$

$\lim_{x \to 0} \frac{2 \sin^2 x}{x^2} = 2 \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 = 2 (1)^2 = 2$.