what is the limit as x-->3 for (2x^2/x-3)-(6x/x-3)...this can also be simplified as what is the limit as x-->3 for (2x^2-6x)/(x-3)?
what is the limit as x--> infinity for (2x^3-3x^2+5x-1)/(4x^3+2x^2-x+10)?....please show work for both answers...thanks a bunch!
2007-01-28 13:41:46 · 3 answers · asked by nyadastar 2 in Science & Mathematics ➔ Mathematics
what is the limit as x-->3 for (2x^2/x-3)-(6x/x-3)...this can also be simplified as what is the limit as x-->3 for (2x^2-6x)/(x-3)?
= 2x(x-3(/(x-3) = 2x, so lim x --> 3 is 2*3 = 6
what is the limit as x--> infinity for (2x^3-3x^2+5x-1)/(4x^3+2x^2-x+...
Sorry but the last part of your problem is missing. I would suggest that you try L'Hospital's rule , when you find what's missing..
2007-01-28 13:51:19 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
1) First question: Factor the numerator so that you have 2x(x - 3)/(x - 3). Then cancel the (x-3) terms and your left with 2x. Lim as x approaches 3 is 2*3 = 6.
2) Second question: Divide numerator and denominator by the highest power of x which is x^3. Then you have 2 - 3/x + 5/x^2 - 1/ x^3 DIVIDED BY 4 +2/x - 1/x^2. As x approaches infinity all the terms with x in the denominator go to zero and your left with 2/4 or 1/2.
2007-01-28 13:57:46 · answer #2 · answered by AARON F 1 · 0⤊ 0⤋
try.. multiplying by the recipricol?
2007-01-28 13:50:42 · answer #3 · answered by SurferDudeJAS 2 · 0⤊ 0⤋