Question

What is the limit of `(x^2 − 5x)/(x^2 − 4x − 5)` as x approaches 5?

Answer 1

`5/6`

Explanation:

`x^2-5x=x(x-5)`
and
`x^2-4x-5=(x-5)(x+1)` thus we get
`(x*(x-5))/((x-5)(x+1))=x/(x+1)` for `xne 5`

Answer 2

`lim_(xrarr5)(x^2 − 5x)/(x^2 − 4x − 5)=lim_(xrarr5)[2x-5]/[2x-4]=[10-5]/[10-4]=5/6`

Explanation:

`lim_(xrarr5)(x^2 − 5x)/(x^2 − 4x − 5)=[25-25]/[25-25]=0/0`

`"L'hospital Rule"`

since the direct compensation product equal `0/0` we will use L'hospital Rule
`color(red)[lim_(trarra)(f'(x))/(g('x))]`

`f(x)=x^2 − 5x`

`f'(x)2x-5`

`g(x)=x^2 − 4x − 5`

`g'(x)=2x-4`

`lim_(xrarr5)(x^2 − 5x)/(x^2 − 4x − 5)=lim_(xrarr5)[2x-5]/[2x-4]=[10-5]/[10-4]=5/6`

Answer 3

`\frac{5}{6}`

Explanation:

You can factor `x` in the numerator:

`x^2-5x = x(x-5)`

You can factor the denominator by finding the roots of the polynomial. In this case, we can use the sum/product method: if the coefficient of `x^2` is `1`, then we can write the equation as `x^2-sx+p`, where `s` is the sum of the solutions and `p` is their product.

So, we're looking for two numbers `x_1` and `x_2` such that:

`x_1+x_2 = 4`
`x_1x_2 = -5`

So, `x_1 = -1` and `x_2 = 5`

And finally `x^2-4x-5=(x+1)(x-5)`

The fraction becomes

`\frac{x^2-5x}{x^2-4x-5}=\frac{x(x-5)}{(x+1)(x-5)}=\frac{x}{x+1}`

So, when `x` approaches `5`, the limit is `\frac{5}{6}`