tolong koreksi ya ka, bener ga jawaban dr nomor 6-10?
Matematika
divadivv231
Pertanyaan
tolong koreksi ya ka, bener ga jawaban dr nomor 6-10?
1 Jawaban
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1. Jawaban whongaliem
no.7 jawabanmu salah harusnya ...
[tex] \lim_{x \to \ 9} \frac{ \sqrt{x} - 3}{ x - 9} = \lim_{x \to \ 9} \frac{ \sqrt{x} - 3}{( \sqrt{x} + 3)( \sqrt{x} - 3)} [/tex]
[tex]= \lim_{x \to \ 9} \frac{1}{ \sqrt{x} + 3} [/tex]
[tex]= \frac{1}{ \sqrt{9} + 3} [/tex]
[tex]= \frac{1}{3 + 3} [/tex]
[tex]= \frac{1}{6} [/tex]
no.8 jawabanmu salah, harusnya ...
[tex] \lim_{x \to \infty} \frac{ 3 x^{2} - 4x + 6}{2 x^{2} + x - 5} = \lim_{x \to \infty} \frac{ x^{2} (3 - \frac{4}{x} + \frac{6}{ x^{2} }) }{ x^{2} (2 + \frac{1}{x} - \frac{5}{ x^{2} }) } [/tex]
[tex]= \lim_{x \to \infty} \frac{3 - \frac{4}{x} + \frac{6}{ x^{2} } }{2 + \frac{1}{ x^{2} } - \frac{5}{ x^{2} } } [/tex]
[tex]= \frac{3 - \frac{4}{\infty} + \frac{6}{ \infty^{2} } }{2 + \frac{1}{\infty} - \frac{5}{ \infty^{2} } } [/tex]
[tex]= \frac{3 - 0 + 0}{2 + 0 - 0} [/tex]
[tex]= \frac{3}{2} [/tex]