Question

How do you evaluate the limit `lim(x^2-9x+20)/(x^2-5x)dx` as `x->5`?

Answer 1

`1/5.`

Explanation:

`"The Reqd. Lim.="lim_(x to 5)(x^2-9x+20)/(x^2-5x),`

`=lim{ul(x^2-5x)-ul(4x+20)}/{x(x-5)},`

`=lim{x(x-5)-4(x-5)}/{x(x-5)},`

`=lim_(x to 5){cancel((x-5))(x-4)}/{xcancel((x-5))}`

`=(5-4)/5.`

`:." The Reqd.Lim.="1/5.`