1. 1. LIMIT OF SEQUENCE imn-ooC= c and limn-oo=0 for p>0. After the two limits are rigorously established in Examples 1.2.2 and 1.2.3, the conclusions based on the two limits become solid 1.1.1 Arithmetic rule Intuitively, if a is close to 3 and y is close to 5, then the arithmetic combinations t+y and ry are close to 3+5=8 and 3.5= 15. The intuition leads to the following property of limit Proposition1.13( Arithmetic rule). Suppose limn→xn= l and limn→∞n=k lim (an +yn)=l+k, lim can=c, lim Ingn=kl, lim En I n→0 where c is a constant and kf0 in the last equality The proposition says limn→(xn+mn)=limn→xn+limn→n. However,the equality is of different nature from the equality in Proposition 1.1.2, because the convergence of the limits on two sides are not equivalent: If the two limits on the ght converge, then the limit on the left also converges and the two sides are equal However, for In=(1)" and n =(1)n+, the limit limn-yoo(an +yn)=0 on the left converges. but both limits on the right diverge Exercise111. Explain that limn→∞n= l if and only if limn→∞(xn-l)=0. Exercise 1. 1.2. Suppose In and yn converge. Explain that limn-yoo nin=0 implies either limn-oo In=0 or limn-+oo n =0. Moreover, explain that the conclusion fails if an and n are not assumed to converge Example 1. 1.2. We have 2+ 2m2+n lim li 2+lin mn→1-limn→∞=+limn+nmn→ lin 2+0 0+0·01.1. LIMIT OF SEQUENCE 9 limn→∞ c = c and limn→∞ 1 np = 0 for p > 0. After the two limits are rigorously established in Examples 1.2.2 and 1.2.3, the conclusions based on the two limits become solid. 1.1.1 Arithmetic Rule Intuitively, if x is close to 3 and y is close to 5, then the arithmetic combinations x+y and xy are close to 3+5 = 8 and 3·5 = 15. The intuition leads to the following property of limit. Proposition 1.1.3 (Arithmetic Rule). Suppose limn→∞ xn = l and limn→∞ yn = k. Then limn→∞ (xn + yn) = l + k, limn→∞ cxn = cl, limn→∞ xnyn = kl, limn→∞ xn yn = l k , where c is a constant and k 6= 0 in the last equality. The proposition says limn→∞(xn + yn) = limn→∞ xn + limn→∞ yn. However, the equality is of different nature from the equality in Proposition 1.1.2, because the convergence of the limits on two sides are not equivalent: If the two limits on the right converge, then the limit on the left also converges and the two sides are equal. However, for xn = (−1)n and yn = (−1)n+1, the limit limn→∞(xn + yn) = 0 on the left converges, but both limits on the right diverge. Exercise 1.1.1. Explain that limn→∞ xn = l if and only if limn→∞(xn − l) = 0. Exercise 1.1.2. Suppose xn and yn converge. Explain that limn→∞ xnyn = 0 implies either limn→∞ xn = 0 or limn→∞ yn = 0. Moreover, explain that the conclusion fails if xn and yn are not assumed to converge. Example 1.1.2. We have limn→∞ 2n 2 + n n2 − n + 1 = limn→∞ 2 + 1 n 1 − 1 n + 1 n2 = limn→∞  2 + 1 n  limn→∞  1 − 1 n + 1 n2  = limn→∞ 2 + limn→∞ 1 n limn→∞ 1 − limn→∞ 1 n + limn→∞ 1 n · limn→∞ 1 n = 2 + 0 1 − 0 + 0 · 0 = 2