1. 1. LIMIT OF SEQUENCE n+a n+c n+b n+d b n2+d n +b td 3+b +b-+d9.+b-边+d Exercise 1.1.9. Find the limits n2+ain+ao n2+cin+ 1 n+b n+d n+ n-+aln+a0 n-+Cin+ co m2+ bin+b0 n2+din+do Exercise 1.1.10. Find the limits, p, q >0 +c +d anP+bn9+c 又.n2p+a1m+a2 anq+bnp+c m2q+61n9+b 1.1.2 Sandwich rule The following property reflects the intuition that if a and z are close to 3, then anything between s and z should also be close to 3 Proposition 1.1.4 (Sandwich Rule). Suppose In Un En for sufficiently big n. If mn→xn=limn→oxn=l, then lim→oyn=l lote that something holds for sufficiently big n is the same as something fails for only finitely many n havinple 1.1.3. By 2n-3 >n for sufficiently big n(in fact, n>3 is enough),we E 0<√2n-3 Then by lin→a0=limn→ 0 and the sandwich rule, we get lim √2n-3 On the other hand, for sufficiently big n, we have n+1 2n and n-1 and therefore vn+I√2n2V21.1. LIMIT OF SEQUENCE 11 1. n n + 1 − n n − 1 . 2. n 2 n + 1 − n 2 n − 1 . 3. n √ n + 1 − n √ n − 1 . 4. n + a n + b − n + c n + d . 5. n 2 + a n + b − n 2 + c n + d . 6. n + a √ n + b − n + c √ n + d . 7. n 3 + a n2 + b − n 3 + c n2 + d . 8. n 2 + a n3 + b − n 2 + c n3 + d . 9. √ n + a √3 n + b − √ n + c √3 n + d . Exercise 1.1.9. Find the limits. 1. n 2 + a1n + a0 n + b − n 2 + c1n + c0 n + d . 2. n 2 + a1n + a0 n2 + b1n + b0 − n 2 + c1n + c0 n2 + d1n + d0 . 3.  n + a n + b 2 −  n + c n + d 2 . 4.  n 2 + a n + b 2 −  n 2 + c n + d 2 . Exercise 1.1.10. Find the limits, p, q > 0. 1. n p + a nq + b . 2. anp + bnq + c anq + bnp + c . 3. n p + a nq + b − n p + c nq + d . 4. n 2p + a1n p + a2 n2q + b1nq + b2 . 1.1.2 Sandwich Rule The following property reflects the intuition that if x and z are close to 3, then anything between x and z should also be close to 3. Proposition 1.1.4 (Sandwich Rule). Suppose xn ≤ yn ≤ zn for sufficiently big n. If limn→∞ xn = limn→∞ zn = l, then limn→∞ yn = l. Note that something holds for sufficiently big n is the same as something fails for only finitely many n. Example 1.1.3. By 2n − 3 > n for sufficiently big n (in fact, n > 3 is enough), we have 0 < 1 √ 2n − 3 < 1 √ n . Then by limn→∞ 0 = limn→∞ 1 √ n = 0 and the sandwich rule, we get limn→∞ 1 √ 2n − 3 = 0. On the other hand, for sufficiently big n, we have n + 1 < 2n and n − 1 > n 2 , and therefore 0 < √ n + 1 n − 1 < √ 2n n 2 = 2 √ 2 √ n