# Download A First Course in Real Analysis (2nd Edition) (Undergraduate by Murray H. Protter, Charles B. Morrey Jr. PDF

By Murray H. Protter, Charles B. Morrey Jr.

Many alterations were made during this moment variation of A First direction in genuine Analysis. the main obvious is the addition of many difficulties and the inclusion of solutions to lots of the odd-numbered routines. The book's clarity has additionally been better via the additional rationalization of a few of the proofs, extra explanatory comments, and clearer notation.

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Extra info for A First Course in Real Analysis (2nd Edition) (Undergraduate Texts in Mathematics)

Example text

Ii) The condition 0 < [x - al < ~ (excluding the possibility x = a) is used rather than the condition [x - al < ~ as in the definition of continuity since J may not be defined at a itself. PROBLEMS In Problems 1 through 8 the functions are continuous at the value a given. In each case find a value ~ corresponding to the given value of e so that the definition of continuity is satisfied. Draw a graph. 1. 01 2. 01 3. 01 4. 1 5. 01 6. f(x) = x 3 - 7. f(x) = x 3 + 3x, a = 8. 2. Limits 35 In Problems 9 through 17 the functions are defined in an interval about the given value of a but not at a.

EXAMPLE 2. Solve for x: 3 x -<5 (x :1= 0). Solution. Since we don't know in advance whether x is positive or negative, we cannot multiply by x unless we impose additional conditions. We therefore separate the problem into two cases: (i) x is positive, and (ii) x is negative. The desired solution set can be written as the union of the sets S, and S2 defined by s, = {x:~ < 5 and x> o}, S2 = {x:~ < 5 and x< o}. Now S, X E <::> 3 < 5x <::> X <::> x>l <::> 3 > 5x <::> X X < O. Similarly, X E S2 x>0 and >!

We assume that L > M and reach a contradiction. Let us define e = (L - M)/2; then from the definition of limit there are positive numbers (jl and (j2 such that and If(x) - LI < e for all x satisfying 0 < Ix - al < (jl Ig(x) - MI < s for all x satisfying 0 < I x - al < (j2. We choose a positive number (j which is smaller than (jl and (j2 and furthermore so small that f(x) ~ g(x) for 0 < Ix - al < (j. In this interval, we have M - e < g(x) < M Since M +e and + s = L - s, it follows that g(x) < M + s = L - e < f(x) < L + e.