Pertanyaan

Tentukan nilai limit berikut. x → ∞ L im ( x − 1 3 x − x + 1 2 x )

Tentukan nilai limit berikut.

$x \to \infty L im (\frac{3 x}{x - 1} - \frac{2 x}{x + 1})$

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

nilai dari adalah 1.

nilai dari $stack L i m with x rightwards arrow infinity below open parentheses fraction numerator 3 x over denominator x minus 1 end fraction minus fraction numerator 2 x over denominator x plus 1 end fraction close parentheses$ adalah 1.

Pembahasan

Jadi, nilai dari adalah 1.

$table attributes columnalign right center left columnspacing 0px end attributes row cell stack L i m with x rightwards arrow infinity below open parentheses fraction numerator 3 x over denominator x minus 1 end fraction minus fraction numerator 2 x over denominator x plus 1 end fraction close parentheses end cell equals cell stack L i m with x rightwards arrow infinity below open parentheses fraction numerator 3 x left parenthesis x plus 1 right parenthesis minus 2 x left parenthesis x minus 1 right parenthesis over denominator left parenthesis x minus 1 right parenthesis left parenthesis x plus 1 right parenthesis end fraction close parentheses end cell row blank equals cell stack L i m with x rightwards arrow infinity below open parentheses fraction numerator 3 x squared plus 3 x minus 2 x squared plus 2 x over denominator x squared minus 1 end fraction close parentheses end cell row blank equals cell stack L i m with x rightwards arrow infinity below open parentheses fraction numerator x squared plus 5 x over denominator x squared minus 1 end fraction close parentheses end cell row blank equals cell stack L i m with x rightwards arrow infinity below open parentheses fraction numerator begin display style x squared over x squared end style plus begin display style fraction numerator 5 x over denominator x squared end fraction end style over denominator begin display style x squared over x squared end style minus begin display style 1 over x squared end style end fraction close parentheses end cell row blank equals cell stack L i m with x rightwards arrow infinity below fraction numerator 1 plus begin display style 5 over x end style over denominator 1 minus begin display style 1 over x squared end style end fraction end cell row blank equals cell fraction numerator 1 plus 0 over denominator 1 minus 0 end fraction end cell row blank equals 1 end table$

Jadi, nilai dari $stack L i m with x rightwards arrow infinity below open parentheses fraction numerator 3 x over denominator x minus 1 end fraction minus fraction numerator 2 x over denominator x plus 1 end fraction close parentheses$ adalah 1.