Question

How do you use the epsilon delta definition to find the limit of `(x+5)/(2x+3)` as x approaches `-1`?

Answer 1

`lim_(x->-1) (x+5)/(2x+3) = 4`

Explanation:

Write `x` as:

`x = -1+xi = xi-1`

and evaluate:

`abs (f(x)-L) = abs ((xi-1 + 5)/(2(xi-1) + 3 )- L) = abs ((xi+4)/(2xi+1) -L) =abs ((xi+4-2xi L -L)/(2xi+1) ) = abs ((xi(1-2L)+(4-L))/(2xi+1) ) <= abs( (xi(1-2L))/(2xi+1) ) +abs ((4-L)/(2xi+1))`

We see that if we set `L=4` the second term vanishes and:

`abs (f(x)-4) <= abs( (-7xi)/(2xi+1) )= (7abs(xi)) /abs(1 +2 xi)`

So for any given `epsilon > 0` we can choose `delta_epsilon > 0` such that

`delta_epsilon < min(1/2, epsilon /7)`

Now for `abs(xi) < delta_epsilon` we have:

`xi > -1/2 => abs (1+2x) > 1`

so that:

`abs (f(x)-4) <= (7abs(xi)) /abs(1 +2 xi) < 7abs(xi)`

`abs(xi) < epsilon/7 => abs (f(x)-4) <= 7abs(xi) < epsilon`

and we have proved that:

`lim_(x->-1) (x+5)/(2x+3) = 4`