Bagi yang ngerti, makasih yang udah mau bantuin
Tentukan nilai limit
Jawaban:
4. 8/5
5.
[tex]2 \sqrt{2} [/tex]
Penjelasan dengan langkah-langkah:
4.
[tex] \lim _{x \to4} \frac{x - 4}{x - \sqrt{3x + 4} } \\ = \lim_{x \to4} \frac{x - 4(x + \sqrt{3x + 4)} }{(x - \sqrt{3x + 4} )(x + \sqrt{3x + 4}) } \\ = \lim_{x \to4} \frac{(x - 4)(x + \sqrt{3x + 4} )}{ {x}^{2} - 3x - 4 } \\ = \lim_{x \to4} \frac{(x - 4)(x + \sqrt{3x + 4}) }{(x - 4)(x + 1)} \\ = \ lim_{x \to4} \frac{x + \sqrt{3x + 4} }{x + 1} \\ = \frac{4 + \sqrt{3(4) + 4} }{4 + 1} \\ = \frac{4 + 4}{5} \\ = \frac{8}{5} [/tex]
5.
[tex] \lim _{x \to2} \frac{x - 2}{ \sqrt{x} - \sqrt{2} } \\ = \lim_{x \to2} \frac{ (x - 2)(\sqrt{x} + \sqrt{2}) }{( \sqrt{x} - \sqrt{2})( \sqrt{x} + \sqrt{2} )} \\ = \lim_{x \to2} \frac{(x - 2)( \sqrt{x} + \sqrt{2}) }{x - 2} \\ = \ log_{x \to2} \sqrt{x} + \sqrt{2} \\ = \sqrt{2} + \sqrt{2} \\ = 2 \sqrt{2} [/tex]
[answer.2.content]
