Question

How do you calculate this limit without using l’Hospital’s rule: lim [ (x+2 / x-3) - ( x^2 -7x / x^2 -2x -3) ] as x → 3 ?

Answer 1

`lim_(x->3) [(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3)]`

does not exist, but:

`lim_(x->3^-) [(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3)] = -oo`

`lim_(x->3^+) [(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3)] = +oo`

Explanation:

Evaluate:

`lim_(x->3) [(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3)]`

Note that:

`x^2-2x-3 = (x+1)(x-3)`

so:

`(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3) = 1/(x-3)( x+2 -(x^2-7x)/(x+1))`

`(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3) = 1/(x-3)( (x+2 )(x+1)-(x^2-7x))/(x+1)`

`(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3) = 1/(x-3)( x^2+3x+2-x^2+7x)/(x+1)`

`(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3) = 1/(x-3)( 10x+2)/(x+1)`

`(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3) = 2/(x-3)( 5x+1)/(x+1)`

Then:

`lim_(x->3) [(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3)] = lim_(x->3) 2/(x-3)( 5x+1)/(x+1)`

and as in `x=3` the second fraction is continuous:

`lim_(x->3) [(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3)] = 2(lim_(x->3) 1/(x-3))(lim_(x->3)( 5x+1)/(x+1))`

`lim_(x->3) [(x+2)/(x-3) - (x^2-7x)/(x^2-2x-3)] = 8lim_(x->3) 1/(x-3)`

But we know that
`lim_(x->3) 1/(x-3)`

does not exist and:

`lim_(x->3^-) 1/(x-3) = -oo`

`lim_(x->3^+) 1/(x-3) = oo`