1 3 k 1 dk lim x 1 x 1 3 k 1 dk 2 u substitution u 3

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1 3 k + 1 dk  =  lim x → ∞ 1 x 1 3 k + 1 dk 2) u -substitution:  u = 3 k +1,  dk=   1 3 du   3) Find new bounds of integral: a. k =1    u =3(1)+1= 4 b. k =x    u =3(x)+1= 3x+1 4)   lim x → ∞ 1 3 4 3 x + 1 1 u du 5) Set  1 3  aside, and evaluate  lim x → ∞ 4 3 x + 1 1 u du  =  ln(u) 6) lim x → ∞ 1 3 4 3 x + 1 1 u du lim x →∞ ln ( u ) 3 ] 4 3 x + 1   7) Evaluate:  lim x → ∞ ln ( | 3 ( x ) + 1 | ) 3 ln ( 4 ) 3 8) Take the limit as x ln ( 3 () ) 3 ln ( 4 ) 3  =  ln ( ) 3
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a. The limit is  ln ( ) 3  because as it goes to  , it is unbounded. The subtraction of  ln ( 4 ) 3  is so minimal that it will not affect the behavior of the function as it  approaches  . D. Since the integral is infinite, ln ( ) 3 , the series is infinite as well. This makes the infinite series divergent. Sources Weir, M. D., Hass, J., & Thomas, G. B. (2014). Thomas calculus: Early transcendentals . Boston: Pearson.
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  • Summer '17
  • lim, Indian mathematics

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