limits - please help

sade · #1 $Old$ September 19th, 2007, 06:28 PM

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

2. lim Öx-2
x®4 x-4

Find the limits algebraically:
3. lim (5x2-9x-8)
t®4

4. lim x-2
x®2 x3-8

5. lim 4-Ö16+h
h®0 h

6. lim ( x2 - 1 )
x® 1 x-1 x-1

Find each limit, if it exists.
7. lim |x-4|
x® 4- x-4

8. lim |x-4|
x® 4+ x-4

9. lim |x-4|
x® 4 x-4

10. lim 1
x® 8- x-8

11. lim 1
x® 8+ x-8

12. lim 1
x®8 x-8

topsquark · #2 $Old$ September 20th, 2007, 07:36 AM

Quote:

Originally Posted by sade $View Post$

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

There are no restrictions on the domain of (3x - 1)5 = 15x - 5, so just plug in x = -1.

Quote:

Originally Posted by sade $View Post$

2. lim Öx-2
x®4 x-4

Is this
$\lim_{x \to 4}\frac{\sqrt{x} - 2}{x - 4}$

If so, then note that $x - 4 = (\sqrt{x} + 2)(\sqrt{x} - 2)$ . Then
$\lim_{x \to 4}\frac{\sqrt{x} - 2}{x - 4} = \lim_{x \to 4}\frac{\sqrt{x} - 2}{(\sqrt{x} + 2)(\sqrt{x} - 2)} = \lim_{x \to 4}\frac{1}{\sqrt{x} + 2}$ .

You take it from here.

-Dan

topsquark · #3 $Old$ September 20th, 2007, 07:39 AM

Quote:

Originally Posted by sade $View Post$

Find each limit, if it exists.
7. lim |x-4|
x® 4- x-4

8. lim |x-4|
x® 4+ x-4

9. lim |x-4|
x® 4 x-4

Using a concept that I just saw someone else use (ThePerfectHacker) in another post, when x approaches 4 from the positive side note that |x - 4| = x - 4. When x approaches 4 from the negative side, |x - 4| = -(x - 4).

If the limits in 7 and 8 exist and are equal, then the limit in 9 exists and is equal to those limits.

-Dan