Evaluate the following limit: `underset( x rarr 5) ( "lim")(x^2-9x+20)/(x^2-6x+5)`

Explanation

Given,

`(lim)_(x->5)(x^2−9x+20)/(x^2−6x+5)`

`=(lim)_(x->5)(5^2−9xx5+20)/(5^2−6xx5+5)`

`=0/0`

which is `0/0` form

So, applying L-Hospital form

`(lim)_(x->5)(x^2−9x+20)/(x^2−6x+5)`

`=(lim)_(x->5)(2x−9)/(2x−6)`

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