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### 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

```

```
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/
```
Associated Topics:
High School Analysis
High School Calculus

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