Spivak Chapter 5 Problem 11/12 Comparing Limits These two problems are quite related and I'm not sure of my answers. *I would appreciate you judging my writing of proofs, English is not my first language and this fall I will start a proof based course at Uni.


*Suppose that $f(x)=g(x)$ when $0<|x-a|<\delta$. Prove that $lim_{x \to a} f(x)=lim_{x \to a} g(x)$


What I did:
For every $\epsilon>0$, there is a $\delta>0$ such that $|f(x)-l|>\epsilon$ whenever $0<|x-a|<\delta$. So, $lim_{x \to a} f(x)=l=lim_{x \to a} g(x)$. 
On the other direction, $lim_{x \to a} f(x)=l=lim_{x \to a} g(x)$ implies $lim_{x \to a} f(x)-g(x)=0$. From that, $|(f(x)-g(x))-0|<\epsilon$ for some $\epsilon>0$, therefore there is a $\delta$ such that $0<|x-a|<\delta$.


*(a) Suppose that $f(x)\leq g(x)$ for all x. Prove that $lim_{x \to a} f(x)\leq\lim_{x \to a} f(x)$, provided these limits exist. Do the same now with only the inequality.


(b) How can these hypotheses be weakened?
What I did:
(a)If $f(x)\leq g(x)$ for all x, then $g(x)-f(x)\geq 0$ for every $x$ in $R$. We get $\lim(g(x)-f(x))\geq 0$, and thus $lim_{x \to a} g(x)\geq\lim_{x \to a} f(x)$
(b) Saying that $f(x)\leq g(x)$ for all $x$ but $x_0$.
Conclusion
So I would be glad if you could help me out with those two questions. Regarding the first one, I'm more in doubt whether the second direction proof is correct and whether I could just repeat the proof without including the equality. 
 A: If you know that $\left|f\left(x\right)-l\right|<\epsilon$ then it immediately follows that $\left|g\left(x\right)-l\right|<\epsilon$.


*

*Assume hypothesis: $lim_{x\rightarrow a}f(x)=l$

*Keep in mind we wish to prove $lim_{x\rightarrow a}g(x)=l$.  To prove a limit equals a value, by and large, you will start by saying "let positive $\epsilon$ be given."  Let positive $\epsilon$ be given.

*By (1), there exists a $\delta_{1}>0$ such that if $0<\left|x-a\right|<\delta_{1}$ then $\left|f\left(x\right)-l\right|<\epsilon$.

*Assume that there exists a $\delta>0$ such that if $0<\left|x-a\right|<\delta$ then $f\left(x\right)=g\left(x\right)$.

*Let $\delta'=\min\left(\delta,\delta_{1}\right)$ and assume that $x$ is a real number such that $0<\left|x-a\right|<\delta'$.

*By virtue of how $\delta'$ is defined, we know that both $0<\left|x-a\right|<\delta_{1}$ and $0<\left|x-a\right|<\delta$.  By (3), $\left|f\left(x\right)-l\right|<\epsilon$.  By (4), $f(x)=g(x)$.  Therefore, $\left|g\left(x\right)-l\right|<\epsilon$.
Wrap up: We start by saying "let   $\epsilon>0$ be given" and end with "there exists a $\delta'>0$ such that if $0<\left|x-a\right|<\delta'$ then $\left|g\left(x\right)-l\right|<\epsilon$.  
That is the definition of $lim_{x\rightarrow a}g(x)=l$.
