Could someone please check my work

Let N E N such that N > n'+ 10 Prove lim 4 0 By the Archimedean property (Theorem 3.3.9), n' 4 10Jim - by dividing each term by n if n 2 N then n 2 N > ( 10 ) and i s = to 10 10 10 6+ SN Ism - 8 = 10 0= 10 10 10 Claim 1. lim (1+ 19) - 0 (V ( : ) Subclaim 1. 1. lim () =0 Thus, lim - ) = 0. Subclaim 1.2. lim - -0 By Theorem 4.2.1(a), lim (7 + 75 ) = lim (" + lim (75 ) = 0 +0 = 0 . Thus, lim ( + 10 ) = 0. Claim 2. lim (6 + -5 =6 WTS lim (6 + 725 ) )= 6 , which by the definition of convergence means WTS lim (= 0, which by the definition of convergence means, VE > 0, 3N E N such that Vn Z N, 6 + 75 - 6 0, 3N E N such that Vn Z N, - - O T =0 -'NZUA 7 5 = = n> - (#) Let N E N such that N >- Let N E N such that N > " (2). By the Archimedean property (Theorem 3.3.9), By the Archimedean Property (Theorem 3.3.9), if n > N then n 2 N > - and _ N then n 2 N > / ( 2 ) and 1 5 . 1 9 9 9 N5 Thus, lim ( ) = 0. Isn - 81 = - 1725 - 0| = 725 5 NB S 9 F E WTS lim - 10 5 = 0, which by the definition of convergence means, Thus , lim (6+ 5 = 6 . VE > 0, AN E N such that Vn > N, JU 75 - O = = n > 10 Therefore, 72 5 n + 10 lim 6n5 + 9 = 0 3. lim 10n + 11n3 9n5 + 2n 0 1 10/9 21/11 0o