Limit ProblemsDate: 08/08/98 at 07:32:35 From: Josephine Subject: Calculus Dear Dr Math, I'm doing a Calculus programme and am having difficulty with some questions on limits. I would appreciate it if you could help. Thanks. (a) lim sqrt(2-t) - sqrt(2) t->0 ------------------- t (b) lim u - sqrt(3) u->srqt(3) ----------- (u^2)-3 (c) lim x^2 - 5*x + 6 x->4 ------------- x^2 - 6*x + 8 (For this question, I tried to factorise both equations first but still could not solve.) (d) lim 3{(x^3) - 1} x->1 ------------ (x-1) Date: 08/08/98 at 08:21:57 From: Doctor Jerry Subject: Re: Calculus Hi Josephine, I'll give you some hints to help get you started on these problems. For problem (a): Try multiplying numerator and denominator by sqrt(2-t) + sqrt(2) before taking the limit. For problem (b): Factor the denominator into (u-sqrt(3))(u+sqrt(3)). For problem (c): The numerator approaches 2 and as x approaches 4 from the right, the denominator, which is (x-4)(x-2), approaches 0 from the left. So the whole fraction goes to -infinity, as x approaches 4 from the right. From the other side, it goes to plus infinity. Thus, the limit doesn't exist. For problem (d): Remember that x^3 - 1 = (x-1)(x^2+x+1). - Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 08/10/98 at 04:52:58 From: Josephine Subject: Re: Calculus Dear Dr. Jerry, I am not able to work out an answer for (a). The answer from the book where I got this question gives sqrt(2)/4. I keep getting infinity. I am not sure if my answer for (d) is correct. I got 7. (The book had a 9 as the answer). Please help. Thanks. Date: 08/10/98 at 08:07:03 From: Doctor Jerry Subject: Re: Calculus Hi Josephine, Here is some more help for those two problems. For problem (a): Multiplying the numerator by sqrt(2-t) + sqrt(2) gives 2-t-2 for the numerator and t(sqrt(2-t) + sqrt(2)) for the denominator. The common factor of t can be removed, leaving: 1/(sqrt(2-t) + sqrt(2)) which goes to 1/(2*sqrt(2)) as t->0. For problem (d): We can cancel (x-1) from numerator and denominator. This leaves 3(x^2+x+1), which approaches 3(3) = 9 as x approaches 1. - Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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