By Serge Lang

This 5th version of Lang's ebook covers all of the subject matters routinely taught within the first-year calculus series. Divided into 5 elements, every one portion of a primary direction IN CALCULUS comprises examples and functions in relation to the subject lined. furthermore, the rear of the booklet includes specified options to numerous the workouts, letting them be used as worked-out examples -- one of many major advancements over past versions.

**Read Online or Download A first course in calculus, Fifth Edition PDF**

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**Extra resources for A first course in calculus, Fifth Edition**

**Example text**

Let be a sequence and let L and M be limits of {a,}El. We need to prove that L = M . 37 we can do so by showing that for all E > 0 the inequality I L - MI < E holds. Let E > 0 be arbitrary but fixed. Then there is an N1 E N such that for all n 2 N1 & we have that la, - L / < -. There also is an N2 E N such that for all n 2 N2 we 2 & have la, - M I < -. Let N := max{Nl, N2]. Because N 2 N1, for all n 2 N we 2 & & have (a, - L / < -. Because N 2 N2, for all n 2 N we have la, - MI < -. 37 we conclude that I L - MI = 0, and hence L = M .

Prove that for all x E d 0 ii. For x E R let f ( x ) := sup { f ( r ) : r E Q and r 5 x ] . Prove that for all x E R the above definition is not self-contradictory by proving it assigns exactly one value to each x E w. ) iii. Prove that the above function does not map any two points to the same image by proving that for all x, y E R the inequality x f y implies that f ( x ) # f ( y ) . ) iv. Prove that the above function “reaches” every element of Jk by proving that for all 1 E there is an x E R so that f ( x ) = 1.

Define the function h : (1, . . , m n } + I by if j 5 n , Then it is easy to show that h is bijective (Exercise h ( j ) := f ( A ; g ( j - n ) ; if j > n. 2-20). But this means that I is finite, a contradiction. 27 A sequence ( u , } ~of~real numbers converges iff it is a Cauchy sequence. Proof. For the direction “jlet , L” := lim a,. We need to prove that for all E > 0 n-oo there is an N E N so that for all m , n 2 N we have \a, - a, I < E . & Let E > 0. Then there is an N E N so that for all n 2 N we have la, - LI < -.