Question

Kuis (medium)
Tentukan hasil dari:

[tex]\rm\displaystyle(a.)~~\lim _{x\to\infty}~~\frac{2x^4}{5x^5-x^2}\\\\(b.)~~\lim _{x\to\infty}~~\frac{4x^5+3x^4~.~.~.}{2x^3-5x^2~.~.~.}\\\\(c.)~~\lim _{x\to\infty}~~6x+3-\sqrt{36x^2+12x+4}\\\\(d.)~~\lim _{x\to\infty}~~\frac{x^5}{x^5}[/tex]

Answer 1

Penjelasan dengan langkah-langkah:

prinsip limit tak hingga = dibagi dengan pangkat terbesar

a. lim x-> tak hingga 2x^4 / 5x^5 - x²

= (2x^4 / x^5) / (5x^5 / x^5) - (x² / x^5)

= 0 / 5 - 0

= 0 / 5 = 0

b. lim x-> tak hingga (4x^5 + 3x^4) / (2x^3 - 5x²)

= ((4x^5 / x^5) + (3x^4 / x^5)) / ((2x^3 / x^5) - (5x² / x^5))

= 4 + 0 / 0 - 0

= 4 / 0

= tak hingga

c. lim x-> tak hingga 6x + 3 - √(36x² + 12x + 4)

= (6x + 3 - √(36x² + 12x + 4) x ((6x + 3) +√(36x² + 12x +4)) / ((6x + 3) +√(36x² + 12x +4))

= (36x² + 36x +9 - 36x² + 12x + 4) / (6x + 3) + √(36x² + 12x + 4)

= (48x + 13) / (6x + 3) + √(36x² + 12x + 4)

= 48 / 6 + √36

= 48 / 6 + 6

= 48 / 12

= 4

d. lim x-> tak hingga x^5 / x^5

= (x^5 / x^5) / (x^5 / x^5)

= 1 / 1

= 1