Math101 Revision - Lecture 14

Math101 – Lecture 14 Revision

We say f(x) à L

As x à c, or        lim f(x) = L

                              x à c

if given ℇ > 0 there exists δ 7 such that |f(x) – L| < ℇ whenever |x à c| < δ

Prove formally that

Lim         (x + 1

____ = 1

1 - x)

X à 0

Proof: Let ℇ > 0 given let |x - 0| < δ, ie x |x| < δ with δ to be determined.

We want to prove à

|x + 1

____ - 1               < ℇ, whenever |x| < δ

1 – x      |

We have to find the δ

x + 1                      1 + x – (1 – x)

____ - 1 =            ___________

1 – x                     1 – x

               =             2x

                              ____

                              1 – x

So,

|x + 1                                  |2x

____ - 1               =             ____

1 – x      |                            1 – x|

                                             =             2|x|                      2δ

                                                            ______  <             ______

                                                            |1 + x|                 |1 + x|

|1 - x| > |1| - |x|

For δ < 1 (so |x| < 1), we have (1 – x)  > 1 - |x| > 1 - |x| > 1 – δ

               1                            1

∴           _______   <            _____

               |1 – x|                  1 – δ

∴

|1 + x                                  2δ                          2δ

_______ - 1        <             _____    <             ____

1 – x           |                       |1 – x|                 1 - δ

                              2δ                                         ℇ

Now choose at _____     = ℇ, ie δ =            ____

                              1 – c                                    2 + ℇ

Then we have

|1 + x                                  2δ          

_______ - 1        <             _____    = ℇ

1 – x           |                       1 – ℇ     

                                             ℇ

Whenever |x| < δ =        ____

                                             2 + ℇ

|x – L| < δ

i.e. – δ < x – c < δ

i.e.

c – δ < x < δ + c

Our limit, so far, are two sided. We take x à c⁺ (approach from above c) and x à c^- (approach from below c)

This will not work for √x as x à 0

We need one sided limits in this case.

eg Discuss the limit x à 1 for

               x², x > 1

f(x) = {

               1 – x, x < 1

Two sided limits at x = 1

From the left x à 1^- , f(x) = 1 – x à 0 but 0 is not in the range of f from the right, x à x⁺ f(x) = x² = 1

Limits as x à + ∞

Are one sided.

x à ∞ is from below.

x à -∞ is from above

Formally: - we say f(x) à L as x à ∞, OR               lim          f(x) = L, if given any ℇ > 0 there exists some x

                                                                                          x à ∞

such that |f(x) - L| < 3, whenever x > X.

Prove that 1/x à 0 as x à ∞

Proof: Let ℇ > 0 be given

               Let x > X, X to be determined

               [We want to prove that |1/x| < whenever x > X]

               We can take x > 0, as x > 0 and |1/x| = 1/x < 1/X

               Choose X = 1/ ℇ (as 1/x = ℇ)

So,  |1/x| < ℇ whenever x > 1/ ℇ             e.g. prove that 1/√(x ²+ 1) à 0 as à ∞

Proof: Let ℇ > 0 be given. Let x > X, x to be determined |1/√(x ²+ 1) – 0| = 1/√(x ²+ 1) < 1/x² <

                                                                                                                        Limit -->

1/X, choose x = 1/ ℇ

Then |1/√(x ²+ 1)| < ℇ for x > X

eg Prove that (√(x + 1) -√(x - 1))/x à as x à ∞

Proof: Let ℇ > 0 be given Let x > X, X to be determined

Firstly (√(x + 1) -√(x - 1))/x           =  ((√(x + 1) -√(x - 1))/x) * ((√(x + 1) + √(x + 1))/(√(x + 1) + √(x + 1))

                                                            = ((x + 1) – (x – 1))/(x(√(x + 1) + √(x + 1))

                                                            = 2/(x(√(x + 1) + √(x + 1)) < 2/x

Since √(x + 1) + √(x + 1) > 1

But 2/x < 2/X, as x > X

∴ (√(x + 1) -√(x - 1))/x < 2/x

Choose x = 2/ℇ (as 2/x = ℇ)

∴ (√(x + 1) -√(x - 1))/x < ℇ

Whenever x > X = 2/ℇ