1) You need to substitute 2 for x in equation under limit such that:

`lim_(x-gt2) (8-x^3)/(x^2-5x+6) = (8-2^3)/(2^2-10+6) `

`lim_(x-gt2) (8-x^3)/(x^2-5x+6) = 0/0`

Since the limit is indeterminate, you should use l'Hospital's limit such that:

`lim_(x-gt2) (8-x^3)/(x^2-5x+6) = lim_(x-gt2) ((8-x^3)')/((x^2-5x+6)') `

`lim_(x-gt2) (8-x^3)/(x^2-5x+6) = lim_(x-gt2) (-3x^2)/(2x-5)`

You need to substitute...

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1) You need to substitute 2 for x in equation under limit such that:

`lim_(x-gt2) (8-x^3)/(x^2-5x+6) = (8-2^3)/(2^2-10+6) `

`lim_(x-gt2) (8-x^3)/(x^2-5x+6) = 0/0`

Since the limit is indeterminate, you should use l'Hospital's limit such that:

`lim_(x-gt2) (8-x^3)/(x^2-5x+6) = lim_(x-gt2) ((8-x^3)')/((x^2-5x+6)') `

`lim_(x-gt2) (8-x^3)/(x^2-5x+6) = lim_(x-gt2) (-3x^2)/(2x-5)`

You need to substitute 2 for x in equation under limit such that:

`lim_(x-gt2) (-3x^2)/(2x-5) = (-3*2^2)/(2*2-5)`

`lim_(x-gt2) (-3x^2)/(2x-5) = -12/(-1)`

`lim_(x-gt2) (-3x^2)/(2x-5) = 12`

**Hence, evaluating the limit yields `lim_(x-gt2) (8-x^3)/(x^2-5x+6) = 12.` **

2) You need to substitute -1 for x in equation under limit such that:

`lim_(x-gt-1) (x^2-5x+6)/(x^2-3x+2) = (1+5+6)/(1+3+2)`

`lim_(x-gt-1) (x^2-5x+6)/(x^2-3x+2) = 12/6`

`lim_(x-gt-1) (x^2-5x+6)/(x^2-3x+2) = 2`

** Hence, evaluating the limit yields `lim_(x-gt-1) (x^2-5x+6)/(x^2-3x+2) = 2.` **

3) You need to substitute 5 for x in equation under limit such that:

`lim_(x-gt5) ((x^2-4x-5)/(x^2-8x+15))-x = ((25-20-5)/(25-40+15))-5 = 0/0 - 5`

Since, the limit `lim_(x-gt5) ((x^2-4x-5)/(x^2-8x+15)) ` yields an idetermination, you may use l'Hospital's theorem such that:

`lim_(x-gt5) (x^2-4x-5)/(x^2-8x+15) = lim_(x-gt5) ((x^2-4x-5)')/((x^2-8x+15)')`

`lim_(x-gt5) (x^2-4x-5)/(x^2-8x+15) = lim_(x-gt5) (2x-4)/(2x-8)`

`lim_(x-gt5) (2(x-2))/(2(x-4)) = lim_(x-gt5) (x-2)/(x-4)`

You need to substitute 5 for x in equation under limit such that:

`lim_(x-gt5) (x-2)/(x-4) = (5-2)/(5-4) = 3`

**Hence, evaluating the limit yields `lim_(x-gt5) ((x^2-4x-5)/(x^2-8x+15))-x= -2` .**

`lim_(x->2) ((8-x^3)/(x^2-5x+6))`

Firt we will plug in x= 2 .

`==gt lim_(x->2) ((8-8)/(4-10+6))= 0/0`

Then, we notcie that x= 2 is a zero both numerator and denominator.

==> Then we will factor and simplify .

`==gt lim_(x->2) (8-x^3)/(x^2-5x+6) `

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

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

`= -(4+4+4)/(2-3) = 12 `

`==gt lim_(x->2) (8-x^3)/(x^2-5x+6) = 12`

``