LIMIT~
[tex]\lim\limits_{x \to 1} \dfrac{\sqrt{2-x}-x}{x^{2}-x}=\cdot[/tex]
Nt: susah ;)
Penjelasan dengan langkah-langkah:
[tex] \displaystyle \:\lim_{x \to \: 1} \frac{ \sqrt{2 - x} - x }{x {}^{2} - x} \\ \\ \displaystyle \:\lim_{x \to \: 1} \frac{ \frac{d}{dx} \sqrt{ 2 - x} - x }{ \frac{d}{dx} x {}^{2} - x} \\ \\ \displaystyle \:\lim_{x \to \: 1} \frac{ \frac{1}{2 \sqrt{2 - x} } \times ( - 1) - \frac{d}{dx}(x) }{2x - \frac{d}{dx}(x) } \\ \\ \displaystyle \:\lim_{x \to \: 1} \frac{ - \frac{1}{2 \sqrt{2 - x} } - 1 }{2x- 1} \\ \\ \displaystyle \:\lim_{x \to \: 1} \frac{ - \big(\frac{1}{2 \sqrt{2 - x} } + 1 \big) }{2x- 1} \\ \\ \displaystyle \:\lim_{x \to \: 1} \frac{ - \big(\frac{1}{2 \sqrt{2 - x} } + \frac{2 \sqrt{2 - x} \times 1 }{2 \sqrt{2 - x} \times 1 } \big) }{2x- 1} \\ \\ \displaystyle \:\lim_{x \to \: 1} \frac{ - \frac{1 + 2 \sqrt{2 - x} }{2 \sqrt{2 - x} } }{2x- 1} \\ \\ \displaystyle \:\lim_{x \to \: 1} - \frac{ \frac{1 + 2 \sqrt{2 - x} }{2 \sqrt{2 - x} } }{ 2x- 1} \\ \\ \displaystyle \:\lim_{x \to \: 1} {- \frac{1 + 2 \sqrt{2 - x} }{2 \sqrt{2 - x} } } \div (2x- 1) \\ \\ \displaystyle \:\lim_{x \to \: 1} {- \frac{1 + 2 \sqrt{2 - x} }{2 \sqrt{2 - x} (2x - 1)}} \\ \\ - { \frac{1 + 2 \sqrt{2 - 1} }{2 \sqrt{2 - 1} (2 \times 1 - 1)}} \\ \\ - { \frac{1 + 2 \sqrt{1} }{2 \sqrt{1} (2 - 1)}} \\ \\ - { \frac{1 + 2 \sqrt{1} }{2 \sqrt{1} \times 1}} \\ \\ - \frac{1 + 2 \times 1}{2 \times 1} \\ \\ - \frac{3}{2} [/tex]
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