Prove the following limit. lim x → 5 3x − 8 = 7 solution 1. preliminary analysis of the problem (guessing a value for δ). let ε be a given positive number. we want to find a number δ such that if 0 < |x − 5| < δ then |(3x − 8) − 7| < ε. but |(3x − 8) − 7| = |3x − 15| = 3 . therefore, we want δ such that if 0 < |x − 5| < δ then 3 < ε that is, if 0 < |x − 5| < δ then < ε 3 . this suggests that we should choose δ = ε/3. 2. proof (showing that δ works). given ε > 0, choose δ = ε/3. if 0 < < δ, then |(3x − 8) − 7| = = 3 < 3δ = 3 = ε. thus if 0 < |x − 5| < δ then |(3x − 8) − 7| < ε. therefore, by the definition of a limit lim x → 5 3x − 8 = 7.

means to say that for any given , we can find such that anytime (i.e. the whenever is "close enough" to 5), we can guarantee that (i.e. the value of is "close enough" to the limit value).

What we want to end up with is

Dividing both sides by 3 gives

which suggests is a sufficient threshold.

The proof itself is essentially the reverse of this analysis: Let be given. Then if

and so the limit is 7. QED

center= (-2,0)

radius= sqrt2/2

step-by-step explanation:

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0