# Proving the limit laws

- Last modification date
- 2016-05-01
- Generated on
- 2019-01-14

**Addition.** Let E⊂R be a subset of the real numbers, f,g:E→R be real functions, and L1,L2∈R be real numbers. Now suppose limx→af(x)=L1 and limx→ag(x)=L2. We would like to show that limx→a(f(x)+g(x))=L1+L2.

Since limx→af(x)=L1 and limx→ag(x)=L2, we know that ∀ϵ>0∃δ>0∀x(0<|x−a|<δ→|f(x)−L1|<ϵ) and ∀ϵ>0∃δ>0∀x(0<|x−a|<δ→|g(x)−L2|<ϵ) Now let ϵ>0. We need to find δ>0 such that 0<|x−a|<δ→|(f(x)+g(x))−(L1+L2)|<ϵ But |f(x)+g(x)−(L1+L2)|=|f(x)−L1+g(x)−L2|≤|f(x)−L1|+|g(x)−L2| by the triangle inequality. So if we can show |f(x)−L1|+|g(x)−L2|<ϵ, then our result follows immediately. But, ϵ2 being a positive real number, we have δ1,δ2>0 such that 0<|x−a|<δ1→|f(x)−L1|<ϵ20<|x−a|<δ2→|f(x)−L2|<ϵ2 So if we let δ:=min(δ1,δ2), then |f(x)−L1| and |g(x)−L2| are both less than ϵ2, so |f(x)−L1|+|g(x)−L2|<ϵ2+ϵ2=ϵ which completes the proof. □