tentukan limit dibawah ini menggunakan cara diferensial
(soal di gambar)
(soal di gambar)
Jawab:
Penjelasan dengan langkah-langkah:
[tex]\displaystyle \lim_{x \to 2} \dfrac{\sqrt{4x+1} - 3}{3x-6} \stackrel{\begin{aligned} 4x+1 = t^2 \\ x =\dfrac{t^2-1}{4} \end{aligned}}{\stackrel{}{=\joinrel=}} \dfrac{1}{3}\lim_{t \to 3} \dfrac{t-3}{\dfrac{t^2-1}{4} - 2} = \dfrac{4}{3}\lim_{t \to 3} \dfrac{t-3}{t^2-9} \\ \lim_{x \to 2} \dfrac{\sqrt{4x+1} - 3}{3x-6} = \dfrac{4}{3}\lim_{t \to 3} \dfrac{1}{2t} \\\\ \boxed{\boxed{\lim_{x \to 2} \dfrac{\sqrt{4x+1} - 3}{3x-6} = \dfrac{4}{3}\cdot \dfrac{1}{2\cdot 3} = \dfrac{2}{9}}}[/tex]
semoga bermanfaaat..............
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