Penjelasan dengan langkah-langkah:
limit
bagian ( a )
[tex] \rm lim_{ \: x \: \to \: 0} \: \frac{3x}{ \sqrt{1 + x} - \sqrt{1 - x} } \\ [/tex]
merasionalkan penyebut bentuk akar
[tex] \rm \frac{3x}{ \sqrt{1 + x} - \sqrt{1 - x} } \: . \: \frac{ \sqrt{1 + x} + \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \\ \rm\frac{3x( \sqrt{1 + x} + \sqrt{1 - x}) }{1 + x - (1 - x)} = \frac{3x( \sqrt{1 + x} + \sqrt{1 - x} )}{2x} \\ \frac{3( \sqrt{1 + x} + \sqrt{1 - x} )}{2} [/tex]
kemudian, substitusi nilai x = 0
[tex] \frac{3( \sqrt{1 + x} + \sqrt{1 - x} )}{2} = \frac{3( \sqrt{1 + 0} + \sqrt{1 - 0} ) }{2} \\ \frac{3( \sqrt{1} + \sqrt{1} )}{2} = \frac{3(2)}{2} = 3[/tex]
bagian ( b )
[tex] \rm lim_ { \: x \: \to \: 0 } \: \frac{6 - \sqrt{9 - 2x} }{x} \\ [/tex]
[tex] \rm \frac{6 - \sqrt{9 - 2x} }{x} \: . \: \frac{6 + \sqrt{9 - 2x} }{6 + \sqrt{9 - 2x} } \\ \rm \frac{ {6}^{2} - (9 - 2x) }{x(6 + \sqrt{9 - 2x} )} = \frac{36 - 9 + 2x}{x(6 + \sqrt{9 - 2x} )} \\ \rm \frac{2x + 27}{x(6 + \sqrt{9 - 2x} )} [/tex]
substitusi nilai x = 0
[tex] \frac{2x + 27}{x(6 + \sqrt{9 - 2x}) } = \frac{2(0) + 27}{0(6 + \sqrt{9 - 2(0)} } \\ \frac{0 + 27}{0(6 + \sqrt{9} )} = \frac{27}{0} = + \infty [/tex]
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Penjelasan dengan langkah-langkah:
limit
bagian ( a )
[tex] \rm lim_{ \: x \: \to \: 0} \: \frac{3x}{ \sqrt{1 + x} - \sqrt{1 - x} } \\ [/tex]
merasionalkan penyebut bentuk akar
[tex] \rm \frac{3x}{ \sqrt{1 + x} - \sqrt{1 - x} } \: . \: \frac{ \sqrt{1 + x} + \sqrt{1 - x} }{ \sqrt{1 + x} + \sqrt{1 - x} } \\ \rm\frac{3x( \sqrt{1 + x} + \sqrt{1 - x}) }{1 + x - (1 - x)} = \frac{3x( \sqrt{1 + x} + \sqrt{1 - x} )}{2x} \\ \frac{3( \sqrt{1 + x} + \sqrt{1 - x} )}{2} [/tex]
kemudian, substitusi nilai x = 0
[tex] \frac{3( \sqrt{1 + x} + \sqrt{1 - x} )}{2} = \frac{3( \sqrt{1 + 0} + \sqrt{1 - 0} ) }{2} \\ \frac{3( \sqrt{1} + \sqrt{1} )}{2} = \frac{3(2)}{2} = 3[/tex]
bagian ( b )
[tex] \rm lim_ { \: x \: \to \: 0 } \: \frac{6 - \sqrt{9 - 2x} }{x} \\ [/tex]
merasionalkan penyebut bentuk akar
[tex] \rm \frac{6 - \sqrt{9 - 2x} }{x} \: . \: \frac{6 + \sqrt{9 - 2x} }{6 + \sqrt{9 - 2x} } \\ \rm \frac{ {6}^{2} - (9 - 2x) }{x(6 + \sqrt{9 - 2x} )} = \frac{36 - 9 + 2x}{x(6 + \sqrt{9 - 2x} )} \\ \rm \frac{2x + 27}{x(6 + \sqrt{9 - 2x} )} [/tex]
substitusi nilai x = 0
[tex] \frac{2x + 27}{x(6 + \sqrt{9 - 2x}) } = \frac{2(0) + 27}{0(6 + \sqrt{9 - 2(0)} } \\ \frac{0 + 27}{0(6 + \sqrt{9} )} = \frac{27}{0} = + \infty [/tex]