we ve got f is a function knowledge on R
f(x) = (1/2)x+3 + [(3x)/(x^2+1)]
Find : lim f(x) ...( for - and + infities)
proof that f (x) = (x^4-4x^2+7)/[2(x^2+1)^2]
(sorry for my crime in english . i don t study math with this language
Solve this pleaase ( need it after 1h) MATHSSS help ..................................?
2016-02-03 2:07 pm
回答 (1)
2016-02-03 2:23 pm
f(x) = (1/2) x + 3 + 3x /(x^2+1)
lim x -->∞ ( (1/2) x + 3 + 3x /(x^2+1) ) = lim x-->∞ (1/2) x + lim x-->∞ 3 + lim x-->∞ 3x /(x^2+1)
lim x-->∞ (1/2) x = ∞
lim x-->∞ 3 = 3
lim x-->∞ 3x /(x^2+1) = 0 (since the numerator has a lower exponent than that of the denominator)
adding, lim x-->∞ (1/2) x + 3 + 3x /(x^2+1) = ∞
lim x-->-∞ ( (1/2) x + 3 + 3x /(x^2+1) ) = lim x-->-∞ (1/2) x + lim x-->-∞ 3 + lim x-->-∞ 3x /(x^2+1)
lim x-->-∞ (1/2) x = -∞
lim x-->∞ 3 = 3
lim x-->-∞ 3x /(x^2+1) = 0 (since the numerator has a lower exponent than that of the denominator)
adding, lim x-->-∞ (1/2) x + 3 + 3x /(x^2+1) = -∞
lim x -->∞ ( (1/2) x + 3 + 3x /(x^2+1) ) = lim x-->∞ (1/2) x + lim x-->∞ 3 + lim x-->∞ 3x /(x^2+1)
lim x-->∞ (1/2) x = ∞
lim x-->∞ 3 = 3
lim x-->∞ 3x /(x^2+1) = 0 (since the numerator has a lower exponent than that of the denominator)
adding, lim x-->∞ (1/2) x + 3 + 3x /(x^2+1) = ∞
lim x-->-∞ ( (1/2) x + 3 + 3x /(x^2+1) ) = lim x-->-∞ (1/2) x + lim x-->-∞ 3 + lim x-->-∞ 3x /(x^2+1)
lim x-->-∞ (1/2) x = -∞
lim x-->∞ 3 = 3
lim x-->-∞ 3x /(x^2+1) = 0 (since the numerator has a lower exponent than that of the denominator)
adding, lim x-->-∞ (1/2) x + 3 + 3x /(x^2+1) = -∞
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