Evaluate the following limit: `underset( x rarr 5) ( "lim")(x^2-7x+12)/(x^2-3x-4)`

Explanation

Given,

`(lim)_(x->5)(x^2−7x+12)/(x^2−3x−4)`

`=(lim)_(x->5)(5^2−7xx5+12)/(5^2−3xx5−4)`

`=0/0`

which is `0/0` form

So, applying L_Hospital Rule,

`(lim)_(x->5)(x^2−7x+12)/(x^2−3x−4)`

`=(lim)_(x->5)(2x−7)/(2x−3)`

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