- #1
mrsmith
- 7
- 0
Homework Statement
Let f and g be functions such that 0≤f(x)≤g(x) for every x near c, except possibility at c. Use the definition of a limit to prove that if lim(x→c)g(x)=0 then lim (x→c)f(x)=0.
This is my attempt at this proof and I'm very poor at it. Can someone check and give me some pointers? Thanks!
2. The attempt at a solution
Definition: Limit lim(x→a)f(x)=L if and only if, given ϵ>0, there exists δ>0 such that 0<|x-a|<δ implies that |f(x)-L|<ϵ.
Given the definition of a limit we can state that lim(x→c)g(x)=0 has the following properties:
0<|x-c|<δ_0 Implies that |g(x)-0|<ϵ_0.
Thus lim(x→c)f(x)=0, 0<|x-c|<δ_1 implies that |f(x)-0|<ϵ_1.
By the original given statement that 0≤f(x)≤g(x), we can argue that ϵ_1< ϵ_0 and δ_1<δ_0
Also by this given since lim(x→c)g(x)=0 and 0≤f(x)≤g(x)
The value of f(x) is between 0≤g(x)
Therefore |g(x)-0|<ϵ_0= |g(x) |<ϵ_0. Thus δ_0= ϵ_0.
Since 0<|x-c|<δ_0 It follows that |g(x)-0|= |g(x) |<δ_0 where δ_0= ϵ_0. Therefore the limit lim(x→c)g(x)=0 holds true. Since this is true for lim(x→c)g(x)=0 it follows that it must also be true for lim(x→c)f(x)=0 because of the given 0≤f(x)≤g(x) properties of the given function and since ϵ_1< ϵ_0 and δ_1<δ_0.