Find limits: 1.) lim x-->2 (8-x^3)/(x^2-5x+6) 2.) lim x-->-1 (x^2-5x+6)/(x^2-3x+2) 3.) lim x-->5 ((x^2-4X-5)/(x^2-8x+15))-x
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calendarEducator since 2011
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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` .
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
`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`
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