1. Basic Limits1.1. Basic Limit Laws1.2. Exercise 2.3.221.3. Exercise 2.3.291.4. Exercise 2.3.301.5. Exercise 2.3.311.6. Exercise 2.3.321.7. Exercise 2.3.38Math 31A 2009.10.01MATH 31A DISCUSSIONJED YANG1. Basic Limits1.1. Basic Limit Laws. Assume that limx→cf(x) and limx→cg(x) exist. Then:(a) Sum Law:limx→c(f(x) + g(x)) = limx→cf(x) + limx→cg(x).(b) Constant Multiple Law: For a ny number k ∈ R,limx→ckf(x) = k limx→cf(x).(c) Product Law:limx→c(f(x)g(x)) =limx→cf(x)limx→cg(x).(d) Quotient Law: If limx→cg(x) 6= 0, thenlimx→cf(x)g(x)=limx→cf(x)limx→cg(x).1.2. Exercise 2.3.22. E valuate the limit limz→1z−1+zz+1.Solution. Recall that limz→1z = 1 and limz→11 = 1. By the Quotient Law,limz→1z−1=limz →11limz →1z=11= 1. B y the Sum Law, limz→1z−1+ z = limz→1z−1+limz→1z = 1 + 1 = 2. By the Sum Law, limz→1z + 1 = 2. So by the Quotient Law,limz→1z−1+zz+1=limz →1z−1+zlimz →1z+1=22= 1. 1.3. Exercise 2.3.29. Can the Quotient Law be applied to evaluate limx→0sin xx?Solution. The Quotient Law requires the limit of the denominator, namely, limx→0x,to exist and be nonzero. This is not the cas e , so we cannot apply directly. 1.4. Exercise 2.3.30. Show that the Product Law canno t be used to evaluatelimx→π/2(x − π/2) tan x.Solution. The Product Law requires the limit o f each factor to exist. However,limx→π/2tan x does not exist. 1.5. Exercise 2.3.31. Give an example where limx→0(f(x) + g (x)) exists but nei-ther limx→0f(x) nor limx→0g(x) exists.Solution. Let f(x) be any function defined on a neighborhood of 0 (but not nec-essarily at 0) such that limx→0f(x) does not exist (e.g., f(x) = 1/x). Let g(x) =−f(x). Then of co urse limx→0g(x) also does not exis t (other wise by the ConstantMultiple Law, limx→0f(x) also exists). But notice f (x) + g(x) is identicaly zeroin a neig hborhood of 0 (but not nece ssarily at 0). So limx→0(f(x) + g (x)) = 0exists. Math 31A Yang 21.6. Exercise 2.3.32. Assume that the limit La= limx→0ax−1xexists and thatlimx→0ax= 1 for all a > 0. Prove that Lab= La+ Lbfor a, b > 0. [Hint:(ab)x− 1 = ax(bx− 1) + (ax− 1).]Solution. By definition, Lab= limx→0(ab)x−1x= limx→0axbx−1x+ax−1x. Sincelimx→0ax= 1 by assumption and limx→0bx−1x= Lbexists by assumption, theProduct Law states limx→0axbx−1x= 1·Lb. Now limx→0ax−1x= Laby assumption,so the Sum Law yields limx→0axbx−1x+ax−1x= Lb+ La. 1.7. Exercise 2.3.38. Assuming that limx→0f(x)x= 1, which of the followingstatements is necessarily true?(a) f(0) = 0.(b) limx→0f(x) = 0.Solution. Remember that the value of f(x) at x = 0 never matters when we evaluatethe limit limx→0f(x). So (a) is not (necessarily) true.Recall that limx→0x = 0, so by the Product Law, limx→0f(x) = limx→0x·f(x)x=limx→0x · limx→0f(x)x= 0 · 1 = 0. Since limx→0f(x)x= 1, and limx→0x = 0, we getlimx→0f(x) = 0.