limit as x approaches 3
x^4-81 / x^3-27
i need to get the answer without using l'hosptial rule thank you to anyone that is willing to help!!!
x^4-81 / x^3-27
i need to get the answer without using l'hosptial rule thank you to anyone that is willing to help!!!
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lim x → 3 (x^4 - 81)/(x^3 - 27)
Note that in the numerator, you have a difference of two squares and in the denominator, a difference of two cubes.
lim x → 3 [(x² - 9)(x² + 9)]/[(x - 3)(x² + 3x + 9)]
Again, a difference of squares in the numerator.
lim x → 3 [(x + 3)(x - 3)(x² + 9)]/[(x - 3)(x² + 3x + 9)]
(x - 3) cancels out.
lim x → 3 [(x + 3)(x² + 9)]/(x² + 3x + 9)
As x goes to 3...
(6 * 18)/(9 + 9 + 9)
= 4
Therefore, the limit is 4.
Note that in the numerator, you have a difference of two squares and in the denominator, a difference of two cubes.
lim x → 3 [(x² - 9)(x² + 9)]/[(x - 3)(x² + 3x + 9)]
Again, a difference of squares in the numerator.
lim x → 3 [(x + 3)(x - 3)(x² + 9)]/[(x - 3)(x² + 3x + 9)]
(x - 3) cancels out.
lim x → 3 [(x + 3)(x² + 9)]/(x² + 3x + 9)
As x goes to 3...
(6 * 18)/(9 + 9 + 9)
= 4
Therefore, the limit is 4.