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[tex] \sf \lim \limits_{x \to 1} \frac{ \sqrt{x} \: - \: 1 }{ \sqrt[3]{x \: + \: 7} \: - \: 2} = ....[/tex]
Penyelesaian Soal :
[tex] \\ [/tex]
[tex]\sf \lim \limits_{x \to 1} \huge\frac{ \sqrt{x} \: - \: 1 }{ \sqrt[3]{x \: + \: 7} \: - \: 2}[/tex]
[tex]\sf \lim \limits_{x \to 1} \huge{(}\frac{ \sqrt{x} \: - \: 1 \times ( \sqrt[3]{x + 7) {}^{2}} + 2 \sqrt[3]{x + 7} + 4}{ x - 1{}}\huge{)}[/tex]
[tex]\sf \lim \limits_{x \to 1} \huge{(}\frac{ \sqrt{x} \: - \: 1 \times ( \sqrt[3]{x + 7) {}^{2}} + 2 \sqrt[3]{x + 7} + 4}{( \sqrt{x} - 1) \times ( \sqrt{x} + 1) {}}\huge{)}[/tex]
[tex]\sf \lim \limits_{x \to 1} \huge{(}\frac{ \cancel{\sqrt{x} \: - \: 1} \times ( \sqrt[3]{x + 7) {}^{2}} + 2 \sqrt[3]{x + 7} + 4}{( \cancel{\sqrt{x} - 1}) \times ( \ \sqrt{x} + 1) {}}\huge{)}[/tex]
[tex]\sf \lim \limits_{x \to 1} \huge{(}\frac{ \sqrt[3]{x + 7) {}^{2}} + 2 \sqrt[3]{x + 7} + 4}{\sqrt{x} + 1 {}}\huge{)}[/tex]
[tex]\sf \huge{(}\frac{ \sqrt[3]{1 + 7) {}^{2}} + 2 \sqrt[3]{1 + 7} + 4}{\sqrt{1} + 1 {}}\huge{)}[/tex]
[tex] \sf \huge \underline{6}[/tex]
[tex] \\ [/tex]
Pelajari Lebih Lanjut :
- Limit fungsi aljabar :
- https://brainly.co.id/tugas/26306413
- Limit fungsi aljabar :
- https://brainly.co.id/tugas/28000419
Detail Jawaban :
- 》Mapel: Matematika
- 》Kelas : 11
- 》Bab : Limit Fungsi Aljabar
- 》Kata Kunci : limit, fungsi, aljabar, akar, sekawan, teorema, l'hospital
[tex] \\ [/tex]
#Let's Learn together brainly.
[tex]\begin{gathered} \\ \end{gathered} [/tex]
[tex]\sf \lim \limits_{x \to 1} \frac{ \sqrt{x} \: - \: 1 }{ \sqrt[3]{x \: + \: 7} \: - \: 2}[/tex]
[tex]\sf \lim \limits_{x \to 1} \frac{x - 1}{ \sqrt[3]{x + 7} - 2} \times \frac{1}{ \sqrt{x} + 1} [/tex]
[tex] \sf12 \times \lim \limits_{x \to 1} \frac{1}{ \sqrt{1} + 1 } [/tex]
[tex] \sf12 \times \frac{1}{2} [/tex]
[tex] \sf = 6[/tex]
[answer.2.content]
