limit x mendekati 1 = x min 1 per akar x kuadrat + 3 min 2

Jawaban:

[tex] \: \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = 2 \\ \\ [/tex]

Penjelasan dengan langkah-langkah:

[tex] \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } \\ \\ [/tex]

[tex] selesaikan \: dengan \: cara \: kali \: bentuk \: sekawan : \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } \times \dfrac{ \sqrt{ {x}^{2} + 3 } + 2}{ \sqrt{ {x}^{2} + 3 } + 2} \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \lim_{x \to \: 1} \dfrac{(x - 1)( \sqrt{ {x}^{2} + 3 } + 2) }{ {x}^{2} + 3 - 4 } \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \lim_{x \to \: 1} \dfrac{(x - 1) \sqrt{ {x}^{2} + 3 } + 2 }{(x - 1)(x + 1)} \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \lim_{x \to \: 1} \dfrac{ \sqrt{ {x}^{2} + 3 } + 2}{x + 1} \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \dfrac{ \sqrt{ {1}^{2} + 3 } + 2}{1 + 1} \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \frac{ \sqrt{4} + 2}{2} \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \dfrac{2 + 2}{2} \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = 2 \\ \\ [/tex]

[tex] selesaikan \: dengan \: aturan \: L'Hopital : \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \lim_{x \to \: 1} \dfrac{ \dfrac{d}{dx}( x - 1)}{ \dfrac{d}{dx}(\sqrt{ {x}^{2} + 3 } - 2) } \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \lim_{x \to \: 1} \dfrac{1}{ \dfrac{2x}{2 \sqrt{ {x}^{2} + 3} } } \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \lim_{x \to \: 1} \dfrac{ \sqrt{ {x}^{2} + 3 } }{x} \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \dfrac{ \sqrt{ {1}^{2} + 3 } }{1} \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = \sqrt{4} \\ \lim_{x \to \: 1} \dfrac{x - 1}{ \sqrt{ {x}^{2} + 3 } - 2 } = 2 \\ \\ [/tex]