A different approach to algebra of limits. I am a beginner who has just begun to learn about differential calculus from last 10 days.Thanks to math stack exchange for giving me the opportunity to post this question.
Now I want you to please have a look at this graph of any real valued function  $y=f(x)$

$$\lim_{x\to c} f(x)=l$$
In this limiting process let $m$ be an abstract least positive quantity by which $x$ can differ with $c$. It means that $x$ is closest to c when it differs with c by a distance $m$.
Similarly, this goes for $n$ too. It means $f(x)$ is closest to limit $l$ when it differs by $n$. $n$ is a positive quantity.
It means that both $m$ and $n$ tend to $0$.
Now take a look at this picture.

I think that these inequalities are intuitively correct as well as new for me.I think all respected mathematicians would already be knowing  about these equations.I did them for 1st time so they are new for me.
Basically my main motive was to prove the algebra of limits which was my concern in my last question too on this website. I did them in this way:-

My seniors told me that my proof is incorrect and is cheated.Even I don't know from where I have cheated.So please inform me if it is incorrect but I say that I have not cheated.
Here, $l_{f}=\lim_{x\to c} f(x)=l$
Similar notations are for others.I think that you are well acquainted with my notations.If I can't convey then comment quickly. I have proved for 1st case only. 2nd can easily be proved.
I have already talked about my proof.Now please help me to feel that what would happen to$m$ if $f'(c)$ is equal to zero.I could not feel this. So I doubt in my these inequalities that whether they are correct.
I tried my best to prove it.I hope that you all would help me out.Well I searched for their proofs which were beyond my scope so I tried to prove them by myself. Please point out anything wrong and give a detailed explanation of the inequalities.
EDIT 1
Hi, I want to say that help me to correct the proof and If we talk about the equality situation in the inequalities then is the proof correct?If it is incorrect then please help me to do the right as well as tell me the situation when we talk about the same for a constant function. Explain me in detail just like the detailing required for 15 year old child. Please edit the math jax portion of this question.Thanks.
EDIT 2
As per beautiful and elaborative discussion on this question as well as an acceptable answer given by Yves Daoust I want to thank everyone. Now I accept that my proof is incorrect but I want to ask here how it can be corrected so that I could approach it in a new manner or suggest me a complete new proof?I hope that I had conveyed my problem properly.Please tell me if there is any issue.I have again edited it because I didn't get the proof in answers but I found mistakes so again I thank everyone.
 A: Your approach doesn't work because


*

*$f'(c)$ requires the computation of a limit, so that the argument is circular,

*the limit may exist even though $f'(c)$ is undefined,

*$\dfrac\epsilon\delta$ has little relation to $f'(c)$ and $\dfrac\epsilon\delta=|f'(c)|$ is false.
A: You should instead have $\mid f(x)-L\mid\le n $ with $L=\lim_{x\to c}f(x)$ meaning that the limiting value is within $n$ units of the actual value at $x=c$ for certain $x$ being within $\epsilon$ of $c$, mathematically, for $\mid x-c\mid \le \epsilon$
So the first claim that $\mid f(x)-L\mid \ge n$ is NOT true, it should be the replaced with $\mid f(x)-L\mid \le n$ for reasons provided above.
Also kindly consider not posting images that contain important parts of the question, it's better to format your work using Mathjax.
A: The very heart of a limit proof is that 
"for every $\epsilon$ you can find a $\delta$", 
where $\epsilon$ (your $n$) is the size of a neighborhood of $L$ and $\delta$ (your $m$) is the size of a neighborhood of $c$, and if $x$ is in this neighborhood of $c$, then $f(x)$ is in this neighborhood of $L$.

Now to prove that the limit of a sum is the sum of the two limits (if they exist), the proof principle goes as follows:


*

*for both functions, for every $\epsilon$ you can find a $\delta$, by hypothesis;

*split $\epsilon$ in two smaller $\epsilon$'s and get the two corresponding $\delta$'s (it doesn't matter if the split is balanced or not);

*if you take the smallest of the two $\delta$'s, then by the triangular equality, the difference $|f(x)+g(x)-L_f-L_g|\le|f(x)-L_x|+|g(x)-L_g|$ remains below $\epsilon$.
Hence, for every $\epsilon$ we can find a $\delta$.
