1. The problem statement, all variables and given/known data 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.