Here we present the proof of assertions 3. and 4. of the following proposition. The proofs of 1. and 2. will be presented in class and here they are assumed. The proofs presented here will not be presented in class.

**Proposition 3.1.4 (Calculus of Limits)**

*Suppose that and are two functions , and that for some we have*

* , and .*

*for some . Then*

*.**If , .**.**If , .**If , and then .*

*Proof of 3:* For a constant function, [Example 3.1.3 (2)]. By 1.

, and

Now, let . Then there is a and a such that,

whenever , and

whenever

Choose . If we then get,

Now

By the Sum Rule (1.):

.

The last line guaranteed by 2.

*Proof of 4: *Let , and assume that . Set

and

By hypothesis there are and such that

and

By the reverse triangle inequality we have

and so if then , and hence

()

Define , so in particular . Now

Thus if , then the inequalities () and () hold so

and

So finally, we have

as required.

The case where is done by setting and and following the same steps (and is left as an exercise)

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May 23, 2011 at 2:08 pm

Jason Fealyjust wondering if there are more marks going for question one than the other questions in ms2001 summer exam?

May 23, 2011 at 4:54 pm

J.P. McCarthyJason,

Embarrassingly enough I can’t remember. I’m under the impression that the split is equal 25-25-25 but I can’t get access to the marking scheme until the morning — it’s kept under lock and key. I will get back to you then.

J.P.

May 24, 2011 at 10:41 am

J.P. McCarthyJason,

All questions carry equal marks.

J.P.

October 5, 2011 at 9:29 am

MS 2001: Week 3 « J.P. McCarthy: Math Page[…] covered from Proposition 1.4.3 to Proposition 2.1.5 inclusive. Here are the proofs of the product and quotient rules for the calculus of […]