Asked by gibbs
Evaluate the following limits:
(I) Lim. (x+1)/[1-sqrt(4+3x)]
X >-1
(II) Lim. (2-3x-5x^2)/(1+2x^2)
X >Infinite
(I) Lim. (x+1)/[1-sqrt(4+3x)]
X >-1
(II) Lim. (2-3x-5x^2)/(1+2x^2)
X >Infinite
Answers
Answered by
Reiny
rationalize the denominator
= (x+1)/[1-√(4+3x)] * (1 + √(4+3x))/(1 + √(4+3x))
= lim (x+1)(1 + √(4+3x))/ (1 - 4 - 3x)
= lim (x+1)(1 + √(4+3x))/ (-3(1+x)
= lim (1 + √(4+3x) )/-3 , as x -->-1
= -2/3
Lim. (2-3x-5x^2)/(1+2x^2) , X >Infinite
as x ---> infinity
the two terms that will "dominate" are -5x^2 on the top and 2x^2 at the bottom, the rest of the terms will become insignificant compared to their value
so we are left with -5x^2/(2x^2 or -5/2
so Lim. (2-3x-5x^2)/(1+2x^2)
X >Infinite
= -5/2
= (x+1)/[1-√(4+3x)] * (1 + √(4+3x))/(1 + √(4+3x))
= lim (x+1)(1 + √(4+3x))/ (1 - 4 - 3x)
= lim (x+1)(1 + √(4+3x))/ (-3(1+x)
= lim (1 + √(4+3x) )/-3 , as x -->-1
= -2/3
Lim. (2-3x-5x^2)/(1+2x^2) , X >Infinite
as x ---> infinity
the two terms that will "dominate" are -5x^2 on the top and 2x^2 at the bottom, the rest of the terms will become insignificant compared to their value
so we are left with -5x^2/(2x^2 or -5/2
so Lim. (2-3x-5x^2)/(1+2x^2)
X >Infinite
= -5/2
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