Limit Problems

Date: 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/