How do you prove limits using the formal definition of a limit?

How can I prove the following limit:





the limit as x goes to infinity of (x/(x-3)) =1





using the formal definition of a limit.|||Given any 蓻 %26gt; 0, we need to find a corresponding positive integer N


such that |x/(x-3) - 1| %26lt; 蓻 for all x %26gt; N.





To do this, note that |x/(x-3) - 1| = 3/(x - 3).


(We can drop the absolute value bars, since we generally want x to be large.


This is certainly the case for x %26gt; 3.)





To find N, solve for x with |x/(x-3) - 1| = 3/(x - 3) %26lt; 蓻


%26lt;==%26gt; (x - 3)/3 %26gt; 1/蓻


%26lt;==%26gt; x %26gt; 3 + 3/蓻.


-----------------------------


Here's the proof:


Given 蓻 %26gt; 0, choose a positive integer N satisfying N %26gt; 3 + 3/蓻.


Then, |x/(x-3) - 1| = 3/(x - 3) %26lt; 3/(N - 3) %26lt; 蓻 for all x %26gt; N, as required.


-----------------------------


I hope this helps!|||some scratch work first:


For any E %26lt; 0 (here "E" means "epsilon") we want |xn - L| %26lt; E


xn = x/(x - 3)


|x/(x + 3) - 1| %26lt; E


|(x - x + 3)/(x - 3)| %26lt; E


|3/(x - 3)| %26lt; E


For any x %26gt; 3, |3/(x - 3)| = 3/(x - 3) %26lt; E


3 %26lt; E(x - 3)


(3 + 3E)/E %26lt; x





Now the actual proof:


Choose any E %26gt; 0. Let n0 = (3 + 3E)/E.


For all n %26gt; n0,


|x/(x - 3) - 1| =


|3/(x - 3)| %26lt;


|3/((3 + 3E)/E - 3)| =


|3/((3 + 3E - 3E)/E)| =


|3/(3/E)| =


E


Therefore,


limit x/(x - 3) = 1


x -%26gt; infinity|||Observe the following:





x / (x-3) = 1 + 3/(x-3).





So, now let's show that the limit as x goes to infinity of 3/(x-3) = 0 (which is the same thing as proving that the entire statement goes to 1.)





It's just a lot of algebra. You can do it.