Real analysis epsilon delta problem

[subsonicman]

I've been reading through Spivak's calculus, and the problem is the answer key i have a hold of is for a different edition so it often doesn't answer the correct questions.

Anyways, here they are:

Chapter 5 problem 10

b. Prove that lim x-> 0 f(x) = lim x-> a f(x-a)
c. Prove that lim x-> 0 f(x) = lim x-> 0 f(x^3)

I'm mainly having a problem with making my solution rigorous.

For b what I have is on the left side we know there is a α_1 for every ε so

|x| < α_1 and |f(x)-m| < ε

We can choose the same epsilon for the right side:

|x-a| < α_2 and |f(x-a)-n| < ε
So the two x's are necessarily related so I replace it in the second equation with y and I also choose α=min(α_1,α_2).

So we have:

|x| < α and |f(x)-m| < ε
|y-a| < α and |f(y-a)-n| < ε

So now I want to say something like, let y-a=x so if ε<|m-n|/2 then it is impossible for both to be fulfilled unless m=n but I'm not sure how to say that rigorously.

I have a similar problem with the next problem.

 

[Office_Shredder]

For b what I have is on the left side we know there is a α_1 for every ε so

|x| < α_1 and |f(x)-m| < ε

We can choose the same epsilon for the right side:

|x-a| < α_2 and |f(x-a)-n| < ε
So the two x's are necessarily related so I replace it in the second equation with y and I also choose α=min(α_1,α_2).

You're making things too complicated at this step. For all \epsilon &gt;0, there is a \delta such that if |x|&lt;\delta, then |f(x)-m|&lt;\epsilon for some m.

Now, if |x-a|&lt;\delta, the above sentence immediately says that |f(x-a)-m|&lt;\epsilon with no additional work