#bijective

http://u.cubeupload.com/Brenorenz/surjectiveinjective1.png

 Bijective

 (2) f: |R -> {0, 1)

 f(x) =      {o, x irrational

              {I, x rational

Not injective

Is surjective

(3)

 http://u.cubeupload.com/Brenorenz/surjectiveinjective2.png

 f: |N à |R

f(x) = x2 – 1

 not surjective

is injective

 -//-

 Limits

Notions f(x) approaches some value  /_ as x approaches some value c € Domain f

              Indicate by f(x) à L as x à c

 Or          lim f(x) = L

              X ->

              The closer I get to x = L the closer x gets to L

 Approximation

              Given a degree of approximation |f(x) – L ) < 0.01

                             Can you find the x that gives this approximation?

                             eg f(x) = x2 à 1 as x à 1

                             Given the approximation |x2 - 1| < 0.01

                             Can we find the possible x’s?

                             Solution |x2 - 1| < 0.01

                             Means -0.01 < x2 – 1 < 0.01

                             Ie 0.99 < x2 < 1.01

                                            ∴ √0.99 < x < √1.01

                                            ∴ |x2 - 1| < 0.01 whenever √0.99 < |x| < √1.01

 Formal Definition of a Limit

We say f(x) à L as x à c or

Lim

xà c

  If given any ℇ > 0 there is a δ > 0 such that |f(x) = L| < ℇ whenever |x - L| < ℇ

Means given the approximation ℇ : - |f(x) - L| < ℇ

We have to find the δ (depend on          ) to make |f(x) - L| < ℇ whenever |x - c| < δ

eg (a) Prove x2 à 1 as x à 1

 Solution: Let ℇ > 0 be given we must find a δ such that |x2 - 1| < ℇ whenever |x – 1| < δ

We must find the approximate δ

Now |x2 - 1|       = |(x - 1)(x + 1)|

                             = |x - 1||x + 1| < |x + 1|

              Also (x - 1) < δ, - δ < x – 1 < δ

              ie 1 – δ < x < 2 + δ

              ∴ 2 – δ < x + 1 < 2 + δ

              Choose δ < 1, then |x + 1| < 3 and 3 δ

              |x2 – 1| < δ |x + 1| < 3 δ Now choose δ ≤ 1/3 ℇ