Some more factoring and limits....?

Question

The professor wen't over this problem in class, but i'm confused about one step of his solution.

Problem : lim ((1/(2x+6))-1/6)/x
x>0

Step 1: lim ((1/(2(x+3))-1/6)/x)
x>0
Step 2: lim (6(x+3) ((1/(2(x+3)-(1/6) / 6x(x+3)
x>0 How does he get 6(x+3) and why?

Step 3: lim ((3-(x+3))/(6x(x+3))
x>0 Then how does he get from step 2 to step 3.... I just don't understand how to get rid of those darn fractions :\

Any help???

Answer 1

... going from
Step 1: ....... to
Step 2:

he has very cleverly multiplied by "1"

1 = 6(x+3) / 6(x+3)

Note that in the "numerator" :
1/(2x+6))-1/6 = 1/2(x+3) - 1/6,
each of the denominators divides "6(x+3)",
making "6(x+3)" the "least common multiple" of

"6" ....... and
"2(x+3)"

this eliminates the ugly fractions in the numerator

6(x+3) times [1/2(x+3) - 1/6] equals

3 - (x + 3)

......... taking care of the denominator
6(x+3) times "x" equals
6x(x+3)

....

tho I don't see what good it does, since as x appoaches zero, the fraction approaches infinity or some such

but the "steps" are valid

Answer 2

This is hard to see because it is a complicated expression written by necessity without the numerator over the denominator.

He is multiplying the numerator and denominator by 6(x + 3).

Step 1:
Limit as x→0 of ({1 / [2(x + 3)] - 1/6} / x)

Multiply numerator and denominator by 6(x + 3).

Step 2:
Limit as x→0 of ({6(x + 3)/[2(x + 3)] - 6(x + 3)/6} / [x*6(x + 3)])

Step 3:
Limit as x→0 of ({3 - (x + 3)} / [6x(x + 3)])

Step 4:
Limit as x→0 of (-x / [6x(x + 3)])

Step 5:
Limit as x→0 of (-1 / [6(x + 3)]) = -1/18

Answer 3

6x(x+3) is the GCD (greatest common denominator). Multiplying top and bottom by the GCD simplifies the addition of fractions.

Going from step 2 to step 3 is multiplying through by the GCD, cancelling factors in each denominator.
So, for the first term, 6(x+3)(1/2(x+3)) = 3. (6(x+3)=3*(2(x+3))
for the second term, 6(x+3) * 1/6 = x+3 (6(x+3)=6 * (x+3)