I am trying to prove the following limit using the delta epsilon definition, $$\lim_{x\to \infty} \frac{x+1}{x+5} =1$$ So I want to prove that $$\forall N>0, \exists \epsilon >0| x >N \rightarrow \frac{x+1}{x+5}-1 < \epsilon$$
I assume I don't need absolute value for $f(x)-L$ since $x \to \infty$.
Now I can do some scratchwork as follows, $$\frac{x+1}{x-5} -1 < \epsilon \rightarrow x > \frac{-4-5\epsilon}{\epsilon}$$ Now I can begin the proof as follows,
Given $\epsilon >0$, choose $N= \frac{-4-5\epsilon}{\epsilon}$
For $x>N$ $$x > \frac{-4-5\epsilon}{\epsilon}\rightarrow ?$$
But I'm not sure where to go from here, even after rearranging to isolate for x. This seems complicated because I have to work my way backwards to the original fraction $\frac{x+1}{x+5} -5 < \epsilon$. Is there anyway I could manipulate this sort of question to To start with the original fraction/limit statement and work towards isolating x and then substituting N? E.g. $$|f(x) -L| = \cdots = cx > cN=c\frac{\epsilon}{c} = \epsilon$$