# Limit Questions

• Jan 17th 2010, 02:03 PM
youmuggles
Limit Questions
Any help would be wonderful, please work out the answers too, I'm just learning!

lim
x→ 2

x^2−4x+4 / x^2+x−6

lim
x→ 0

x / |x|

lim
x→ 1

x^3−1 / x−1

lim
x→ 1

x^2+4x−5 / x^2+x−2

lim
x→ 2

x^3−8 / 12(x−2)
• Jan 17th 2010, 02:11 PM
SubZero
Quote:

Originally Posted by youmuggles
Any help would be wonderful, please work out the answers too, I'm just learning!

lim
x→ 2

x^2−4x+4 / x^2+x−6

i'll start you off

(x-2)(x-2)/(x+3)(x-2)= (x-2)/(x+3)

lim
x→ 2
(x-2)/(x+3) =0
• Jan 17th 2010, 02:20 PM
General
Quote:

Originally Posted by youmuggles
Any help would be wonderful, please work out the answers too, I'm just learning!

lim
x→ 2

x^2−4x+4 / x^2+x−6

lim
x→ 0

x / |x|

lim
x→ 1

x^3−1 / x−1

lim
x→ 1

x^2+4x−5 / x^2+x−2

lim
x→ 2

x^3−8 / 12(x−2)

For the first one:
$\displaystyle \frac{x^2-4x+4}{x^2+x-6}=\frac{(x-2)(x-2)}{(x+3)(x-2)}$
The bad guys cancel, and you will find your limit by direct substitution.
Similary for the rest expect the second one.

For the second one, you should consider the limit as x goes to zero from right and from left.
and remember that:
If $\displaystyle \lim_{x \to a^{+}} f(x) \neq \lim_{x \to a^{-}} f(x)$
then $\displaystyle \lim_{x \to a} f(x)$ does not exist.
• Jan 17th 2010, 02:40 PM
youmuggles
When I do these questions, these are my answers... Could someone please tell me what i am doing wrong..?

Quote:

Originally Posted by youmuggles
Any help would be wonderful, please work out the answers too, I'm just learning!

lim
x→ 2

x^2−4x+4 / x^2+x−6

(x-2)(x-2)
(x-2)(x+3)

(x-2)
(x+3)

((2)-2)
((2+3)

0
5

= 0

Quote:

Originally Posted by youmuggles
lim
x→ 0

x / |x|

x
x

= 0? DNE?

I still don't fully understand the concept with this one..

Quote:

Originally Posted by youmuggles
lim
x→ 1

x^3−1 / x−1

(x-1)(x^2+2x+1)
(x-1)

x^2-2x+1

(1)^2 - 2(1) + 1

1-2+1

= 0

Quote:

Originally Posted by youmuggles
lim
x→ 1

x^2+4x−5 / x^2+x−2

(x+5)(x-1)
(x-1)(x-1)

(x+5)
(x-1)

((1)+5)
((1)-1)

6
0

= DNE

Quote:

Originally Posted by youmuggles
lim
x→ 2

x^3−8 / 12(x−2)

(x-2)(x^2+4x+4)

12(x-2)

(x^2+4x+4)
12

((2)^2 + 4(2) + 4
12

4+8+4
12

=4/3
• Jan 17th 2010, 02:53 PM
General
Quote:

Originally Posted by youmuggles
When I do these questions, these are my answers... Could someone please tell me what i am doing wrong..?

(x-2)(x-2)
(x-2)(x+3)

(x-2)
(x+3)

((2)-2)
((2+3)

0
5

= 0

correct

x
x

= 0? DNE?

I still don't fully understand the concept with this one..

if $\displaystyle x \rightarrow 0^{+}$, then $\displaystyle |x|=x$
if $\displaystyle x \rightarrow 0^{-}$, then $\displaystyle |x|=-x$

(x-1)(x^2+2x+1)
(x-1)
Wrong, This is the formula $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$
Try to do it again.
x^2-2x+1

(1)^2 - 2(1) + 1

1-2+1

= 0

(x+5)(x-1)
(x-1)(x-1)
No No No, Try to factor the denominator again.
(x+5)
(x-1)

((1)+5)
((1)-1)

6
0

= DNE

(x-2)(x^2+4x+4)
Same mistake, The formula is $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$
Try to do it again.
12(x-2)

(x^2+4x+4)
12

((2)^2 + 4(2) + 4
12

4+8+4
12

=4/3

.................
• Jan 17th 2010, 03:19 PM
youmuggles
(x-2)(x^2+2x+4)
12(x-2)

(x^2+2x+4)
12

((2)^2 + 2(2) + 4
12

4+4+4
12

=1
• Jan 17th 2010, 03:25 PM
General
Quote:

Originally Posted by youmuggles
(x-2)(x^2+2x+4)
12(x-2)

(x^2+2x+4)
12

((2)^2 + 2(2) + 4
12

4+4+4
12

=1

Mwahahaha !
Yes ! You got it !
You > Calculus (Clapping)
You are correct.
Try to use LaTeX to write your math equations.
No need for this complicated way.