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Tentukan nilai limit berikut. x→ +4muylim​(x3−xy2x4−y4​)

Pertanyaan

Tentukan nilai limit berikut.

limit as x rightwards arrow thin space y of open parentheses fraction numerator x to the power of 4 minus y to the power of 4 over denominator x cubed minus x y squared end fraction close parentheses  

S. Yoga

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

 limit as x rightwards arrow thin space y of open parentheses fraction numerator x to the power of 4 minus y to the power of 4 over denominator x cubed minus x y squared end fraction close parentheses equals 2 y.

Pembahasan

Mencari nilai limit:

limit as x rightwards arrow thin space y of open parentheses fraction numerator x to the power of 4 minus y to the power of 4 over denominator x cubed minus x y squared end fraction close parentheses equals limit as x rightwards arrow thin space y of open parentheses fraction numerator open parentheses x squared close parentheses squared minus open parentheses y squared close parentheses squared over denominator x cubed minus x y squared end fraction close parentheses equals limit as x rightwards arrow thin space y of open parentheses fraction numerator open parentheses x squared plus y squared close parentheses open parentheses x squared minus y squared close parentheses over denominator x cubed minus x y squared end fraction close parentheses equals limit as x rightwards arrow thin space y of open parentheses fraction numerator open parentheses x squared plus y squared close parentheses open parentheses x plus y close parentheses open parentheses x minus y close parentheses over denominator x open parentheses x squared minus y squared close parentheses end fraction close parentheses equals limit as x rightwards arrow thin space y of open parentheses fraction numerator open parentheses x squared plus y squared close parentheses open parentheses x plus y close parentheses open parentheses x minus y close parentheses over denominator x open parentheses x plus y close parentheses open parentheses x minus y close parentheses end fraction close parentheses equals limit as x rightwards arrow thin space y of open parentheses fraction numerator open parentheses x squared plus y squared close parentheses open parentheses x minus y close parentheses over denominator x open parentheses x minus y close parentheses end fraction close parentheses equals limit as x rightwards arrow thin space y of open parentheses fraction numerator x squared plus y squared over denominator x end fraction close parentheses equals fraction numerator y squared plus y squared over denominator y end fraction equals fraction numerator 2 y squared over denominator y end fraction equals 2 y 

Jadi, limit as x rightwards arrow thin space y of open parentheses fraction numerator x to the power of 4 minus y to the power of 4 over denominator x cubed minus x y squared end fraction close parentheses equals 2 y.

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Pertanyaan serupa

Tentukan nilai limit berikut dengan menggunakan substitusi langsung x→3lim​4x+25x−1​

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Jawaban terverifikasi

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