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MATH 140 (141)
Ewa Duma (13)
Lecture 8

Lecture 8 - 2.3.pdf

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Department
Mathematics & Statistics (Sci)
Course
MATH 140
Professor
Ewa Duma
Semester
Summer

Description
7. lim c ▯ c 8. lim x ▯ a xla xla These limits are obvious from an intuitive point of view (state them in words or draw graphs of y ▯ c and y ▯ x ), but proofs based on the precise definition are requested in the exercises for Section 2.4. If we now putngf▯x▯ ▯ x:in Law 6 and use Law 8, we get another useful special limit. Apart from guessing using a table how to get the limit, we can also use limit laws to 9. lim x ▯ a n where n is a positive integer xcalculate a limit. This is the method we will need to use in order to solve the problems on the exam. A siSuppose that c is a constant and the limits: lim (x > a) f(x) AND lim (x > a) g(x) exist, cise 37then:ection 2.4.) 10. lim s x ▯ (sa> a) [where+ngis a positive integerx) + lim (x > a) g(x) xla • lim (x > a) [f(x) - g(x)] = lim (x > a) f(x) - lim (x > a) g(x) (Ifn •slim (x > a) [cf(x)] = c lim (x > a) f(x) • lim (x > a) [f(x)g(x)] = lim (x > a) f(x) ⋅ lim (x > a) g(x) • lim (x > a) [f(x) / g(x)] = lim (x > a) f(x) / lim (x > a) g(x)… if lim (x > a) g(x) ≠ 0 More generally, we have the following law, which is proved as a consequence of Law 10 in Section 2.5. To understand these limit laws, we can verbally interpret them as the following: • The limit of a sum is the sum of the limits. ROOT LAW 11. lim sf▯x) ▯ nlim f▯x) where n is a positive integer xla • The limstxla a difference is the difference of the limits. • The limit of a constant times a function is the constant times the limit of the Ifn ifunction.e assume thatlim f▯x▯ ▯ 0. [ xla ] • The limit of a product is the product of the limits. • The limit of a quotient is the quotient of the limits (provided that the limit of the denominator is not 0). EXAMPLE 2 Evaluate the following limits and justify each step. Example: Prof uses example from textbookx ▯ 2x ▯ 1 (a) lim ▯2x ▯ 3x ▯ 4▯ (b) lim xl5 xl▯2 5 ▯ 3x Evaluate the following limit and justify each step: lim (x > 5) (2x^2 - 3x + 4) SOLUTION (a) lim ▯2x ▯ 3x ▯ 4▯ ▯ lim ▯2x ▯ ▯ lim ▯3x▯ ▯ lim 4 (by Laws 2 and 1) xl5 xl5 xl5 xl5 2 ▯ 2 lxl5x ▯ 3 lixl5 ▯ limxl5 (by 3) 2 ▯ 2▯5 ▯ ▯ 3▯5▯ ▯ 4 (by 9, 8, and 7) ▯ 39 Understand that you can use a direct substitution property whereby if f is a polynomial or a rational function and “a” is in the domain of f, then: lim (x > a) f(x) = f(a) We can say that functions with the direct substitution property are called continuous at “a”. and Archimedes never explicitly formulated the concept of a limit. Likewise, mathematicians Functions with the Direct Substitution Property are called continuous at a and will be such as Cavalieri, Fermat, and Barrow, the imme- diate precursors of Newton in the developmentdied in Section 2.5. However, not all limits can be evaluated by direct substitution, as the following examples show. of calculus, did not actually use limits. It was Isaac Newton who was the first to talk explicitly about limits. He explained that the main idea x ▯ 1 behind limits is that quantities “approach nearer 3 Find lim . xl1 x ▯ 1 than by any given difference.” Newton stated Example: Find lim (x > 1) [(x^2-1)/(x-1)] that the limit was the basic concept in caSOLUTION Let f▯x▯ ▯ ▯x ▯ 1▯▯▯x ▯ 1▯ . We can’t find the limit by substituting x ▯ 1 but it was left to later mathematicians like Cauchy to clarify his ideas about lim
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