determine limits using algebraic manipulation

Question

determine limits using algebraic manipulation

beginning in the first cell, find the answer. search for that answer elsewhere in the document, mark that cell #2 and work the problem in that box. process in this manner until you complete the circuit. show all pertinent work. note: this assignment contains two separate circuits.

#1 lim 12 answer: -3/8 # lim (2x-2)/(x-1) answer:0 # if f(x)={x-1, x≤3 find lim f(x) answer:3/8

x->5 x->1 {2x-3, x>3 x->3^-

# lim (s²+6s+5)/(s²-3s-4) answer:2 # lim(3x²-4x-1) answer:8 # lim [(1/2)+(1/t)]/(2+t) answer:1/4

s->-1 x->5 t->-2

# lim[(√x+4)-2]/x answer:-1 # lim4x answer:12 # lim (x²+4x+4)/x² answer:54

x->0 x->2 x->-2

# if f(x)={cos x-sin pi, x≤pi, find lim f(x) answer: dne # lim (x³+x²-4x-4)/(x⁴-16) answer: -4/5

{x-pi-1, x>pi x->pi x->2

# lim (5x²-2x³+2x-5)/(4x²+6x-10x³) answer:-1/4 #1 lim pi answer:-1 # lim (x²-16)/(x-4) answer: 4/3

x->1 x->0 x->4

# lim (x²-4x+4)/(x²+x-6) answer: 12 # lim(2x-4)/(x-1) answer: 11 # lim (5x³-7x²+2^x-2) answer:pi

x->2 x->4 x->1

# lim [(1/x-1)+(1/x²-3x+2)] answer: 2√3 # lim (h²)/[(√h²+h+3)-√h+3)] answer:1/4

x->1 h->0

# lim[(√x²+x+4)-2]/(x²+x) answer: dne # lim (t³+8)/(t+2) answer:8 # lim(2x³-54)/(x-3) answer:0

x->0 t->2 x->3

# if f(x)={x²+x, x≤-1, find lim f(x) answer: 54 # lim (3y⁴-6y³-2y) answer: -2

{-2^-x, x>-1 x->-1 y->-1

Answer 1

The first thing to understand about a circuit worksheet is what those printed answers actually are, because that trips up almost everyone. The number printed inside a box is not the answer to that box. It is the answer to some other box, and it is sitting there so that whoever solves that other problem knows to come here next. So you work your starting box, get a number, then hunt for the box with that number printed in it, and that is your box #2. Once you see that, the whole page stops looking random.

That means the actual work is just evaluating all of them, so here is every limit on your page worked out. I did each one symbolically and then checked it again numerically by closing in on the point from both sides, so these are solid.

lim(x→5) 12                              = 12     constant, nothing to do
lim(x→1) (2x−2)/(x−1)                    = 2      factor the 2 out, cancel (x−1)
lim(x→3−) f(x), f = x−1 on x≤3           = 2      left side, so use the x≤3 branch
lim(s→−1) (s²+6s+5)/(s²−3s−4)          = −4/5   (s+1)(s+5)/((s+1)(s−4))
lim(x→5) (3x²−4x−1)                   = 54     just substitute
lim(t→−2) [(1/2)+(1/t)]/(2+t)          = −1/4   combine the top to (t+2)/(2t), cancel
lim(x→0) [√(x+4)−2]/x                = 1/4    multiply by the conjugate
lim(x→2) 4x                              = 8      substitute
lim(x→−2) (x²+4x+4)/x²                 = 0      top is (x+2)², bottom is 4, not indeterminate
lim(x→π) f(x), cos x − sinπ / x−π−1     = −1     see the note below
lim(x→2) (x³+x²−4x−4)/(x⁴−16)         = 3/8
lim(x→1) (5x²−2x³+2x−5)/(4x²+6x−10x³)  = −3/8
lim(x→0) π                               = π      constant
lim(x→4) (x²−16)/(x−4)                  = 8      (x−4)(x+4)/(x−4)
lim(x→2) (x²−4x+4)/(x²+x−6)            = 0      (x−2)²/((x−2)(x+3)) → (x−2)/(x+3)
lim(x→4) (2x−4)/(x−1)                   = 4/3    substitute, nothing cancels
lim(x→1) (5x³−7x²+2x−2)                = −2     substitute
lim(x→1) 1/(x−1) + 1/(x²−3x+2)          = −1     combine over (x−1)(x−2), it collapses to 1/(x−2)
lim(h→0) h²/[√(h²+h+3) − √(h+3)]      = 2√3   conjugate; the h² cancels exactly
lim(x→0) [√(x²+x+4)−2]/(x²+x)          = 1/4    conjugate; (x²+x) cancels
lim(t→2) (t³+8)/(t+2)                     = 4      see the note below
lim(x→3) (2x³−54)/(x−3)                  = 54     2(x−3)(x²+3x+9)/(x−3)
lim(x→−1) f(x), x²+x on x≤−1 / −2−x on x>−1 = does not exist
lim(y→−1) (3y⁴−6y³−2y)                 = 11     substitute

Remember, the pattern to notice across all of these is that whenever you get 0/0 there is always something to cancel, and there are only three tools that do the cancelling: factor it, combine the fractions on top into one fraction, or multiply by the conjugate when you see a square root. If you can spot which of the three a problem is asking for, the rest is bookkeeping.

Now the honest part, because three things on your page do not line up and you should check them against your paper copy before you build the circuit.

First, the piecewise one at x→π. The left branch gives cosπ − sinπ = −1 − 0 = −1, and the right branch gives π − π − 1 = −1 too. Both sides agree, so that limit does exist and equals −1. The one that genuinely does not exist is the other piecewise, at x→−1, where the left side gives 0 and the right side gives −1.

Second, (t³+8)/(t+2) as written sends t→2, which is just 16/4 = 4 with no cancelling at all. No other box on the page points to 4, which is what makes me think it is meant to be t→−2. That is the 0/0 version, it factors as (t+2)(t²−2t+4)/(t+2), and it gives 12, which does have a home on the page.

Third, two different boxes come out to 2, the (2x−2)/(x−1) one and the piecewise at x→3 from the left, but only one box on the page has 2 printed in it. Two boxes cannot both point to the same next box, so one of those two is copied down slightly differently from your original.

Once you confirm those against your worksheet, chaining the circuit is quick: start at either box marked #1, find its value in the list above, and go to the box with that number printed in it.