Simplifying Limits

[Dazed]

Hi, I am having some trouble simplifying these limits. Any help would be greatly appreciated. Thanks

1. lim (9-x) / lx-9l
x-->9


2. lim sin^(3)4x / sin5xtan^(2)3x
x-->0


3. lim x^(2)+x-12 / 20-13x+2x^(2)
x--> infinity

 

[tkhunny]

The First One

Split it up, 9+ and 9-. What can you do with the absolute values if you KNOW what the sign of the expression is?

The Last One

Multiply numerator and denominator by (1/x^2) and simplify separately.

The Middle One

Sorry, I have to go and it looks like that one might take longer.

 

[Dazed]

umm

thanks for some help but I'm still struggling....

 

[soroban]

Hello, Dazed!

#2 is pretty awful, but we can bust it if we know this theorem:

. . . . . . . sin θ
. . . lim . ------- . = . 1
. . .θ->0 . . θ

. . . . . . . . . . sin³(4x)
2) . lim . --------------------
. . x->0 . sin(5x) tan²(3x)
.

. . . . . . . . . .(sin 4x)³ . . . .1 . . . . cos²3x
We have: . ------------ . --------- . ----------
. . . . . . . . . . . . 1 . . . . . sin 5x . . sin²3x


I'll multiply each fraction by a special "one":

. . . (4x)³ . (sin 4x)³. .5x . . . .1 . . . . (3x)² . . . . 1
. . . ------ . ----------- . ---- . --------- . -------- . ----------- . cos²3x
. . . (4x)³ . . . . 1 . . . . 5x . .sin 5x . . (3x)² . .(sin 3x)²


Then we have:

. . . .64x³ . (sin 4x)³ . .1 . . . 5x . . . .1 . . . .(3x)²
. . . ------- . ----------- . --- . -------- . ----- . ----------- . cos²3x
. . . . .1 . . . . (4x)³ . . .5x . .sin 5x . 9x² . (sin 3x)²

. . . 64x³ . 1 . . .1. . . . . . . . .(sin 4x )³(. . 5x . ) (. . 3x . )²
. . . ------ . --- . ---- . cos²3x (---------) (---------) (---------)
. . . . .1 . . 5x . 9x² . . . . . . . ( . 4x . .) .( sin 5x). ( sin 3x)


The first three fractions reduce to 64/45.

. . As x→0, cos²(3x) → 1

. . And as x→0, the last three fractions also → 1.

Therefore, the limit is 64/45

 

[Dazed]

aahhh

thank you for your help soroban !! Could you also help me with the 1st and 3rd problems too?

 

[soroban]

Hello, Dazed!

tkhunny already threw you some big hints.
Evidently, you didn't catch them . . .

1)
lim (9-x) / lx-9l
x->9

The limit doesn't exist . . . but I suppose we have to show our work.

Since we have an absolute value, we must consider the two cases:
. . . (1) the inside quantity is positive
. . . (2) the inside quantity is negative.

<u>Case 1</u>: .x - 9 is positive (that is, x > 9).
. . . . . . .Then |x - 9| .= .x - 9

. . . . . . . . . . . . . . 9 - x . . . .-(x - 9)
The function is: . ------- . = . -------- . = . -1
. . . . . . . . . . . . . . x - 9 . . . . . x - 9

Therefore: . lim (-1) . = . -1
. . . . . . . . . x->9


<u>Case 2</u>: . x - 9 is negative. (That is, x < 9.)
. . . . . . Then: .|x - 9| .= .9 - x

. . . . . . . . . . . . . . 9 - x
The function is: . ------- . = . 1
. . . . . . . . . . . . . . 9 - x

Therefore: . lim (1) . = . 1
. . . . . . . . . x->9


Since the two values are <u>not</u> equal, the limit does not exist.

3)
lim (x^2 + x - 12)/(20 - 13x + 2x^2)
x -> oo

Divide top and bottom by x².

. . . . . . . . . . . . . .1 + 1/x - 12/x²
We have: . lim . ---------------------
. . . . . . . x->oo . 20/x² - 13/x + 2

As x → oo (infinity), those fractions → 0.

. . . . . . . . . . . . . 1 + 0 - 0 . . . . 1
So the limit is: . ------------ . = . --
. . . . . . . . . . . . . 0 - 0 + 2 . . . . 2