Limit Questions

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January 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)
January 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
January 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:
$\frac{x^2-4x+4}{x^2+x-6}=\frac{(x-2)(x-2)}{(x+3)(x-2)}$
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 $\lim_{x \to a^{+}} f(x) \neq \lim_{x \to a^{-}} f(x)$
then $\lim_{x \to a} f(x)$ does not exist.
January 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
January 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 $x \rightarrow 0^{+}$ , then $|x|=x$
if $x \rightarrow 0^{-}$ , then $|x|=-x$

(x-1)(x^2+2x+1)
(x-1)
Wrong, This is the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$
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 $a^3-b^3=(a-b)(a^2+ab+b^2)$
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

.................
January 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
January 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.