L'Hopital's Rule: Identifying and Solving Indeterminate Forms, Study notes of Mathematics

Download L'Hopital's Rule: Identifying and Solving Indeterminate Forms and more Study notes Mathematics in PDF only on Docsity! Lecture 8: L’Hopital’s rule • Recognize indeterminate forms. Especially, 0/0,∞/∞. Be able to reduce limits of the form 0 · ∞, 10 and 1∞ to evaluating a limit of a quotient. • State L’Hopital’s rule for quotients. • Compute limits using L’Hopital’s rule. Some well-known limits Recall some familiar limits: lim x→0 sin(2x) x = 2 lim x→0 ln(1 + x) x = 1 lim x→0 1− cos(x) x2 = 1 2 The first two may be viewed as difference quotients and this allows us to know the limit. For example, sin(2x) x = sin(2x)− 0 x− 0 and thus with f(x) = sin(2x), lim x→0 sin(2x) x = f ′(0) = 2 Indeterminate forms Each of the limits above can be thought of as lim x→a f(x) g(x) where lim x→a f(x) = lim x→a g(x) = 0. This is an indeterminate form of type 0 0 . As the above examples show, these limits can have many values depending on the functions f and g. This is why the expression f(x)/g(x) is called an indeterminate form. Exercise. Given a value a, can you choose f and g so that the indeterminate form has the value a? If in the above limit we have lim x→a f(x) =∞ and lim x→a g(x) =∞ then this is an indeterminate form of type ∞∞ . (We say the same if one or both of the limits is −∞. Example. lim x→∞ ex x . Indeterminate forms of type 0∞. If lim x→a f(x) = 0 and lim x→∞ g(x) =∞ then lim x→a f(x)g(x) is an indeterminate form. This can be rewritten as 0/0 by considering f(x)/(g(x)−1). Exercise. Can you also rewrite this indeterminate form as ∞/infty? Indeterminate forms of 1∞. We rewrite f(x)g(x) = eg(x) ln(f(x)). If f approaches 1, then ln f approaches 0 and we assume that g has a limit of ∞ Thus on the right-hand side, the exponent is the indeterminate form ∞ · 0. Example. Compute lim x→0 (1 + x)1/x. Solution. We rewrite this as lim x→0 e 1 x ln(1+x) since we evaluated the limit of the exponent above and the exponential function is continuous, we have lim x→0 (1 + x)1/x = e.