• Sep 24th 2007, 06:19 PM
Find the limits using first principle:
1. lim (3x-1)5 (NB 5 is subscript)
x® -1

2. lim Öx-2 (NB this is the square root of x -2)
x®4 x-4

Find the limits algebraically:
4. lim x-2
x®2 x3-8 (NB this is x cubed -8)

5.lim ( x2 - 1 ) (NB this is x squared)
x® 1 x-1 x-1

6.lim 1
x® 8- x-8

7. lim 1
x® 8+ x-8

8. lim 1
x®8 x-8
• Sep 24th 2007, 11:05 PM
CaptainBlack
Quote:

Find the limits using first principle:
1. lim (3x-1)5 (NB 5 is subscript)
x® -1

I will assume you mean:

$
\lim_{x \to -1} (3x-1)^5
$

(the 5 is a super-script).

As $(3x-1)^5$ is continuous at $x=-1$ this limit is just $(3x-1)^5$ evaluated at $x=-1$.

RonL
• Sep 24th 2007, 11:10 PM
CaptainBlack
Quote:

2. lim Öx-2 (NB this is the square root of x -2)
x®4 x-4

As $\sqrt{x-2}$ is continuous at $x=4$ this limit goes to $\infty$.

Now if you mean:

$
\frac{\sqrt{x}-2}{x-4}
$

we have a different problem which you solve by writting: $x-4=(\sqrt{x}-2)(\sqrt{x}+2)$ for $x \ge 0$

RonL
• Sep 25th 2007, 05:26 AM
topsquark
Quote:

Find the limits using first principle:
1. lim (3x-1)5 (NB 5 is subscript)
x® -1

2. lim Öx-2 (NB this is the square root of x -2)
x®4 x-4

Find the limits algebraically:
4. lim x-2
x®2 x3-8 (NB this is x cubed -8)

5.lim ( x2 - 1 ) (NB this is x squared)
x® 1 x-1 x-1

6.lim 1
x® 8- x-8

7. lim 1
x® 8+ x-8

8. lim 1
x®8 x-8

:mad: Did you have questions about when I helped you earlier? It would have been polite to ask me in that post, rather than double posting, which is against the rules.

-Dan
• Sep 25th 2007, 07:09 AM