Matematika

Pertanyaan

[tex] \lim_{x \to \infty} ( \sqrt{x^2 + x (2a+2b) + 4ab} - x)[/tex]

1 Jawaban

  • [tex](\sqrt{x^{2}+x(2a+2b)+4ab}-x)*\frac{\sqrt{x^2{2}+x(2a+2b)+4ab}+x}{\sqrt{x^{2}+x(2a+2b)+4ab}+x}[/tex]
    [tex]=\frac{x^{2}+x(2a+2b)+4ab-x^{2}}{\sqrt{x^{2}+x(2a+2b)+4ab}+x}[/tex]
    [tex]=\frac{(x(2a+2b)+4ab)\frac{1}{x}}{\sqrt{\frac{x^{2}}{x^{2}}+\frac{x(2a+2b)}{x^{2}}+\frac{4ab}{x^{2}}}+\frac{x}{x}}}[/tex]
    [tex]=\frac{(2a+2b)+\frac{4ab}{x}}{\sqrt{1+\frac{2a+2b}{x}+\frac{4ab}{x^{2}}}+1}[/tex]
    masukkin limit x menuju takhingga
    [tex]=\frac{2a+2b}{\sqrt{1}+1}[/tex]
    [tex]=a+b[/tex]

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