Calculus I -Alen Osancliol Quiz Questions Answers limits of function
x -> 3 =5^2.3^3
=25.27 = 675
2-lim (x^3-1)/x^3-2x+4 = (-1)^3-1/(-1)^3-2(-1)+4
x->-1 =-2/5
3-lim (1 + √x)/(√x+4) = (1 + √0)/(√0+4)
x->0(+) =1/2
4-lim sin(πx/2) = sin(π/2)
x -> 1 = 1
5-lim (√x+1)-(√x^2+4)/(x+2)^2-(x+1)^2 = (√0+1)-(√0^2+4)/(0+2)^2-(0+1)^2
x->0 = (1-2)/(4-1)
= -1/3
6-lim(x.tanx) = (π/4).tan(π/4)
x->1 =π/4
7- lim(tan^2x)/(1+cos(π/4)) = [tan^2(π/4)]/[1 +cos(π/4)]
x ->π/4 = (1^2)/1+(√2/2) =1/[(2 +√2)/2] = 2 /(2 +√2)'denominator with 2-√2)
=(4-2√2)/2
=2-√2
8-lim (x^2-1)/ (x-1) = when equation is 0/0 you must be simplification
x -> 1 = ( 1^2-1)/(1-1)
=[(x-1)(x+1)]/(x-1) = x+1
= 1+1
= 2
9- lim (x^3-1)/(x-1) = (1^3-1)/(1-1)
x ->1 = when equation is 0/0 you must be simplification
= x^2 + x +1
= 1^2 + 1 +1
=3
10-lim(x-4)/(√x-2) = (4-4)/[(√4)-2]
x ->4 = when equation is 0/0 you must be simplification
= (√x-2) (√x+2) /(√x-2) = (√x+2)
= (√4+2)
=4
11- lim [(√x+3) -(√3)]/x = [(√0+3) -(√3)]
x ->0 = when equation is 0/0 you must be change the equation if equation not change answer is undefined
12-lim[IxI] = [-x]
x->1(-) = -1
13-lim[IxI] = [x] = 5
x -> -5(+)
14- lim (cosx +1)/sinx = 2/0
x ->0 = (cosx +1)/sinx ' deminator with (cosx-1)'
= cos^2(x)-1/(cosx +1)sinx
= sin^2(x)/(cosx +1)sinx 'simplification'
= sinx/ (cosx +1)
= 0/1 = 0
15-lim (x-1)/[(3√x+7)-2]= (1-1)/(3√1+7)-2)
x ->1 = when equation is 0/0 you must be simplification
= x+7=t^3 x+7=t^3 t^3-7=x t^3-8/t-2 = (t-2)/t^2+2t+4)
=t^2+2t+4
x=t^3-7 1=t^3-7 t=2 4+4+4= 12
16-lim(x +2)/(4√x+18)-2 =(x +2)/(4√x+18)-2
x -> -2 = when equation is 0/0 you must be simplification
= (x +2)/(4√x+18)+2 'deminator with (4√x+18)+2 '
= (x+2)(4√x+18)+2)/(√x+18)-4 'deminator with (√x+18)+4'
=(x+2)[(4√x+18)+2][√x+18)+4]/(x+2)
=[(4√x+18)+2][√x+18)+4]
=[(4√-2+18)+2][√-2+18)+4]
=(2+2)(4+4)
=4.8=32
17-lim(x.sin(1/x)) = 0.sin(1/0)
x ->0 =undefined
18- lim(√x.cos(1/x^2)) = √0.cos[1/0^2]
x ->0 =0
19-lim[(3√x)-1]/[(√x)-1] =
x->1
20-lim[ I2x-1I -I2x+2I/X] = [-2x+1 -2x-1/x]
x ->0 = -4
21- lim[IxI/x] = undefined
x ->0
22-lim(x^2-9)/x^2+2x-3 = (x-3)(x+3)/(x+3)(x-1)
x ->1(+) = -2/0(+)
= -∞
23-lim[(√x+6) - √x]/(x^3)-3(x^2) =[(√x+6) - √x]/x^3-3(x^2)' deminator with(√x+6)-x'
x ->3 = (x+ 6 - x^2)/x^2(x-3)
= -(x-3)(x+2)/x^2(x-3)
= -(x+2)/x^2.(√x+6) - √x)
=-5/54
give back for typos
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