1. Sekolah Dasar
  2. Sekolah Menengah Pertama
  3. Sekolah Menengah Atas
  4. Mata Pelajaran
  5. Programming

Hi Kakak yg sedang baca pertanyan aku skrng...mohon bantu jawab

Berikut ini adalah pertanyaan dari wahyunazihulwan41 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Hi Kakak yg sedang baca pertanyan aku skrng...mohon bantu jawab dong kak :")) soalnyaa tentang nilai limit.Soalnya aku sisipkan di bawah kak....mohon bantu jawab yaa kak :)
Hi Kakak yg sedang baca pertanyan aku skrng...mohon bantu jawab dong kak :

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawaban:

1. \displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}

Kita akan rasionalkan akarnya:

\displaystyle =\lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2} \cdot \frac{\sqrt{x-1}+1}{\sqrt{x-1}+1}

\displaystyle =\lim_{x\rightarrow 2} \frac{x-1-1}{(x-2)(\sqrt{x-1}+1)}

\displaystyle =\lim_{x\rightarrow 2} \frac{x-2}{(x-2)(\sqrt{x-1}+1)}

\displaystyle =\lim_{x\rightarrow 2} \frac{1}{\sqrt{x-1}+1}

Kita kerjakan limitnya seperti biasa (substitusikan x=2)

\displaystyle =\frac{1}{\sqrt{2-1}+1}

\displaystyle =\frac{1}{1+1}

\displaystyle =\frac{1}{2}

Maka:

\displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}\displaystyle =\frac{1}{2}

2. \displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} }

Seperti nomor 1, kita rasionalkan akarnya:

\displaystyle= \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x}}\cdot \frac{1+\sqrt{4+x}}{1+\sqrt{4+x}}

\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{1-(4+x)}}

\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}

\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}

\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-(x+3)}}

\displaystyle = \lim_{x\rightarrow -3} -\left(1+\sqrt{4+x}\right)

\displaystyle = -\left(1+\sqrt{4-3}\right)

\displaystyle = -\left(1+1\right)

\displaystyle = -2

Maka:

\displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} } = -2

3. \displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}

Rasionalkan akarnya:

=\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1}

=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x^2-1)(\sqrt{x}+1)}

=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x+1)(x-1)(\sqrt{x}+1)}

=\displaystyle \lim_{x\rightarrow 1} \frac{1}{(x+1)(\sqrt{x}+1)}

=\displaystyle \frac{1}{(2)(2)}

=\displaystyle \frac{1}{4}

Maka:

\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}=\displaystyle \frac{1}{4}

Jawaban:1. [tex]\displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}[/tex]Kita akan rasionalkan akarnya:[tex]\displaystyle =\lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2} \cdot \frac{\sqrt{x-1}+1}{\sqrt{x-1}+1}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{x-1-1}{(x-2)(\sqrt{x-1}+1)}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{x-2}{(x-2)(\sqrt{x-1}+1)}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{1}{\sqrt{x-1}+1}[/tex]Kita kerjakan limitnya seperti biasa (substitusikan x=2)[tex]\displaystyle =\frac{1}{\sqrt{2-1}+1}[/tex][tex]\displaystyle =\frac{1}{1+1}[/tex][tex]\displaystyle =\frac{1}{2}[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}\displaystyle =\frac{1}{2}[/tex]2. [tex]\displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} }[/tex]Seperti nomor 1, kita rasionalkan akarnya:[tex]\displaystyle= \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x}}\cdot \frac{1+\sqrt{4+x}}{1+\sqrt{4+x}}[/tex][tex]\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{1-(4+x)}}[/tex][tex]\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-(x+3)}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} -\left(1+\sqrt{4+x}\right)[/tex][tex]\displaystyle = -\left(1+\sqrt{4-3}\right)[/tex][tex]\displaystyle = -\left(1+1\right)[/tex][tex]\displaystyle = -2[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} } = -2[/tex]3. [tex]\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}[/tex]Rasionalkan akarnya:[tex]=\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x^2-1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x+1)(x-1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{1}{(x+1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \frac{1}{(2)(2)}[/tex][tex]=\displaystyle \frac{1}{4}[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}=\displaystyle \frac{1}{4}[/tex]Jawaban:1. [tex]\displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}[/tex]Kita akan rasionalkan akarnya:[tex]\displaystyle =\lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2} \cdot \frac{\sqrt{x-1}+1}{\sqrt{x-1}+1}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{x-1-1}{(x-2)(\sqrt{x-1}+1)}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{x-2}{(x-2)(\sqrt{x-1}+1)}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{1}{\sqrt{x-1}+1}[/tex]Kita kerjakan limitnya seperti biasa (substitusikan x=2)[tex]\displaystyle =\frac{1}{\sqrt{2-1}+1}[/tex][tex]\displaystyle =\frac{1}{1+1}[/tex][tex]\displaystyle =\frac{1}{2}[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}\displaystyle =\frac{1}{2}[/tex]2. [tex]\displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} }[/tex]Seperti nomor 1, kita rasionalkan akarnya:[tex]\displaystyle= \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x}}\cdot \frac{1+\sqrt{4+x}}{1+\sqrt{4+x}}[/tex][tex]\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{1-(4+x)}}[/tex][tex]\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-(x+3)}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} -\left(1+\sqrt{4+x}\right)[/tex][tex]\displaystyle = -\left(1+\sqrt{4-3}\right)[/tex][tex]\displaystyle = -\left(1+1\right)[/tex][tex]\displaystyle = -2[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} } = -2[/tex]3. [tex]\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}[/tex]Rasionalkan akarnya:[tex]=\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x^2-1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x+1)(x-1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{1}{(x+1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \frac{1}{(2)(2)}[/tex][tex]=\displaystyle \frac{1}{4}[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}=\displaystyle \frac{1}{4}[/tex]Jawaban:1. [tex]\displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}[/tex]Kita akan rasionalkan akarnya:[tex]\displaystyle =\lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2} \cdot \frac{\sqrt{x-1}+1}{\sqrt{x-1}+1}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{x-1-1}{(x-2)(\sqrt{x-1}+1)}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{x-2}{(x-2)(\sqrt{x-1}+1)}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{1}{\sqrt{x-1}+1}[/tex]Kita kerjakan limitnya seperti biasa (substitusikan x=2)[tex]\displaystyle =\frac{1}{\sqrt{2-1}+1}[/tex][tex]\displaystyle =\frac{1}{1+1}[/tex][tex]\displaystyle =\frac{1}{2}[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}\displaystyle =\frac{1}{2}[/tex]2. [tex]\displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} }[/tex]Seperti nomor 1, kita rasionalkan akarnya:[tex]\displaystyle= \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x}}\cdot \frac{1+\sqrt{4+x}}{1+\sqrt{4+x}}[/tex][tex]\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{1-(4+x)}}[/tex][tex]\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-(x+3)}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} -\left(1+\sqrt{4+x}\right)[/tex][tex]\displaystyle = -\left(1+\sqrt{4-3}\right)[/tex][tex]\displaystyle = -\left(1+1\right)[/tex][tex]\displaystyle = -2[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} } = -2[/tex]3. [tex]\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}[/tex]Rasionalkan akarnya:[tex]=\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x^2-1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x+1)(x-1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{1}{(x+1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \frac{1}{(2)(2)}[/tex][tex]=\displaystyle \frac{1}{4}[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}=\displaystyle \frac{1}{4}[/tex]Jawaban:1. [tex]\displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}[/tex]Kita akan rasionalkan akarnya:[tex]\displaystyle =\lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2} \cdot \frac{\sqrt{x-1}+1}{\sqrt{x-1}+1}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{x-1-1}{(x-2)(\sqrt{x-1}+1)}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{x-2}{(x-2)(\sqrt{x-1}+1)}[/tex][tex]\displaystyle =\lim_{x\rightarrow 2} \frac{1}{\sqrt{x-1}+1}[/tex]Kita kerjakan limitnya seperti biasa (substitusikan x=2)[tex]\displaystyle =\frac{1}{\sqrt{2-1}+1}[/tex][tex]\displaystyle =\frac{1}{1+1}[/tex][tex]\displaystyle =\frac{1}{2}[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow 2} \frac{\sqrt{x-1}-1}{x-2}\displaystyle =\frac{1}{2}[/tex]2. [tex]\displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} }[/tex]Seperti nomor 1, kita rasionalkan akarnya:[tex]\displaystyle= \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x}}\cdot \frac{1+\sqrt{4+x}}{1+\sqrt{4+x}}[/tex][tex]\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{1-(4+x)}}[/tex][tex]\displaystyle =\lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-3-x}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} \frac{(x+3)(1+\sqrt{4+x})}{-(x+3)}}[/tex][tex]\displaystyle = \lim_{x\rightarrow -3} -\left(1+\sqrt{4+x}\right)[/tex][tex]\displaystyle = -\left(1+\sqrt{4-3}\right)[/tex][tex]\displaystyle = -\left(1+1\right)[/tex][tex]\displaystyle = -2[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow -3} \frac{x+3}{1-\sqrt{4+x} } = -2[/tex]3. [tex]\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}[/tex]Rasionalkan akarnya:[tex]=\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}\cdot \frac{\sqrt{x}+1}{\sqrt{x}+1}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x^2-1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{x-1}{(x+1)(x-1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \lim_{x\rightarrow 1} \frac{1}{(x+1)(\sqrt{x}+1)}[/tex][tex]=\displaystyle \frac{1}{(2)(2)}[/tex][tex]=\displaystyle \frac{1}{4}[/tex]Maka:[tex]\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x}-1}{x^2-1}=\displaystyle \frac{1}{4}[/tex]

Semoga dengan pertanyaan yang sudah terjawab oleh Tomaten dapat membantu memudahkan mengerjakan soal, tugas dan PR sekolah kalian.

Apabila terdapat kesalahan dalam mengerjakan soal, silahkan koreksi jawaban dengan mengirimkan email ke yomemimo.com melalui halaman Contact

Last Update: Mon, 23 Aug 21
<< Previous Post
Next Post >>