Jawabanmu

2014-03-27T23:25:58+07:00
√4.2+1 = √9 = 3 - 3 = 0 x² = 4-4=0 sin 0/0 = 0 sin 0 = 0 Jawabannya 0. Semoga bener ♥
limit kan tdk boleh 0/0 jadinya tdk didefinisikan
2014-03-28T00:56:28+07:00
 \lim_{x \to 2}  \frac{ \sqrt{4x+1} -3 }{ x^{2} -4} .\frac{ \sqrt{4x+1} +3 }{  \sqrt{4x+1} +3}

\lim_{x \to 2} \frac{ 4x+1 - 9 }{ (x^{2} -4 )(\sqrt{4x+1} +3)} =\frac{ 4x - 8}{ (x+2)(x - 2 )(\sqrt{4x+1} +3)}= \frac{ 4(x - 2)}{ (x+2)(x - 2 )(\sqrt{4x+1} +3)}

\lim_{x \to 2} \frac{ 4 }{ (x+2 )(\sqrt{4x+1} +3)}  =  \frac{ 4 }{ (2+2 )(\sqrt{9} +3)} =  \frac{1}{6}
itu kan isi sin di soalnya... sinnya dibawa kemna?