Math 151, Spring 2010,cBenjamin Aurispa4.8 Indeterminate Forms and L’Hospital’s RuleWe have seen limits in the past that take the form00,∞∞, and ∞ − ∞. When we encountered these, we hadto do something else...algebra, simplification, factoring...to be able to find the limit. These types of limitsare examples of indeterminate forms.If limx→af(x)g(x)=00or∞∞, then we can use L’Hospital’s Rule to find the limit.L’Hospital’s Rule: Suppose f and g are differentiable functions. If limx→af(x)g(x)=00or∞∞, thenlimx→af(x)g(x)= limx→af0(x)g0(x)Notes: The limit could also be of the form−∞−∞,−∞∞, or∞−∞.If limx→af(x)g(x)=0∞, this is NOT indeterminate: The limit is 0.If limx→af(x)g(x)=∞0, this is NOT indeterminate: The limit will be ∞ or −∞.(1) (#5, 4.8) limx→0ex− 1sin 3x(2) (#14, 4.8) limx→∞(ln x)3x21Math 151, Spring 2010,cBenjamin AurispaIndeterminate Products: If limx→af(x)g(x) = 0 · ∞, this limit is indeterminate. Why?limx→∞1x· x2limx→∞3x2· x limx→∞7x2· x2To find the limit, the goal is to write the indeterminate product in the form00or∞∞and use L’Hospital’sRule.(1) limx→0+csc x ln(1 + sin 7x)(2) (#40, 4.8) limx→−∞xex2Math 151, Spring 2010,cBenjamin AurispaIndeterminate Difference: If limx→a[f(x) − g(x)] = ∞ − ∞, this limit is indeterminate. To find the limit, thegoal is once again to convert this difference into a quotient that we can use L’Hospital’s Rule on if necessary.limx→0+2x + 1sin x−1xIndeterminate Powers: If limx→a[f(x)]g(x)is of the form 00, ∞0, or 1∞, these are indeterminate. These casesare treated by first taking the natural logarithm, which will make the limit of the form 0 · ∞. Then, proceedas we did with indeterminate products. However, we must remember to “undo” the natural logarithm tofind our final answer. (Note that 0∞is NOT an indeterminate form. A limit of this form will be 0.)(1) limx→∞1 +4x2x23Math 151, Spring 2010,cBenjamin Aurispa(2) limx→0+xtan x(3) (#62, 4.8) limx→∞(ex+ x)1/xSummary: There are 7 basic indeterminate forms:00,±∞±∞, ∞ − ∞, 0 · ∞, 00, ∞0,
View Full Document