Heres another problem that I haven't been able to figure out, I need some assistance:
Suppose there is a such that when:
Prove that:
Sounds like a very good teacher!
I once taught a course called "Quantitative Methods for Economics and Business Administration" using a text book assigned by the Business Administration department. On one page of this book we were given three laws of limits:
1) If and then
2) If and then
3) if and and M is not 0, then
On the very next page, they introduced the derivative as the limit of the difference quotient: , completely ignoring the fact that the denominator necessarily goes to 0 so this limit could NEVER be done using only those laws!
Yes, that property is very important, and fairly easy to prove.
Let . Then, by definition of limit, given any there exist such that if then . Note the "<" symbol at the left of |x-a|. It does not matter what happens at x= a because then |x-a|= 0! (I am using the subscript on to distinguish if from the given in the problem.)
Now, given any , we can use the smaller of and to prove that . If 0< |x- a|< the smaller of and , then both and f(x)= g(x) are true. So we can replace f(x) by g(x) in that expression: and we are done!
If the limit for f is equal to L, then:
There exists a delta>0 such that: 0<|x-a|<delta => |f(x)-L|<epsilon. since f(x)=g(x) => |g(x)-L|<epsilon.
So we have: if 0<|x-a|<delta => |g(x)-L|<epsilon. Therefore the limit of g as x approaches a is L. So the limits at a are equal.