# What is the limit: `lim_(x->1)(x^2+2x-3)/(x^2-x)`?

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The limit `lim_(x->1)(x^2+2x-3)/(x^2-x)` has to be determined.

If x is substituted with 1, `(x^2+2x-3)/(x^2-x) = (1+2-3)/(1-1) = 0/0` which is an indeterminate form. l'Hopital's rule can be used here and the numerator and denominator substituted by their derivatives.

This gives: `lim_(x->1)(2x+2)/(2x-1)`

Substituting x = 1 gives `(2 + 2)/(2 - 1) = 4/1 = 4`

**The limit **`lim_(x->1)(x^2+2x-3)/(x^2-x) = 4`

The value of limit `lim_(x->1)(x^2+2x-3)/(x^2-x)` is required.

Substituting x = 1 in `(x^2+2x-3)/(x^2-x)` gives `(1+2-3)/(1-1) = 0/0` which is indeterminate.

To determine the limit, instead of using l'Hospital's rule we can factorize the numerator and denominator. Cancel the common factor and substitute x = 1.

The numerator x^2 + 2x - 3

= x^2 + 3x - x - 3

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

= (x - 1)(x + 3)

The denominator x^2 - x

= x(x - 1)

The given limit `lim_(x->1)(x^2+2x-3)/(x^2-x)`

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

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

Now substitute x = 1. This gives `(1 + 3)/1 = 4/1 = 4`

The limit `lim_(x->1)(x^2+2x-3)/(x^2-x) = 4`

A limit will tell you what the values of y are doing as x approaches a number. In this problem x approaches 1 from both sides. If one is substituted into the denominator you get a zero, which would make this spot undefined, creating an asymptote on your graph. However, you can still find the limit if you find the derivative of both the numerator and the denominator.

x^2-2x+3 = 2x+2

x^2-x = 2x-1

Now you can substitute one into the equation and your limit should be 4.

2(1)+2/2(1)-1 = 4

When your limit cannot be plugged into the "x" value because the limit becomes either `0/0` or `oo/oo` , you may use L'Hospital's Rule. The rule states that the derivative of the top and bottom will be the limit.

Find the derivative of the numerator:

`x^2 + 2x -3 -> 2x +2`

Then find the derivative of the denominator:

`x^2 - x -> 2x - 1`

Your new limit should look like this:

`lim_(x->1) (2x+2)/(2x -1)`

Plug in your limit:

`(2(1) + 2)/(2(1)-1) = 4`

And your limit should be 4!