Principles of Mathematics in Operations Research

206 Solutions

Problems of Chapter 1

l.l

(a) Since, / is continuous at x:

Vei > 0 3Ji > 0 9 Vy 9 \x - y\ < ^ =» |/(x) - f(y)\ < ex.

g is continuous at x:

Ve 2 > 0 382 > 0 3 \/y 9 |x - y\ < S 2 => \g(x) - g(y)\ < e 2.

Fix ei and e 2 at |.

3tfi > 0 9 My 3 \x - y\ < 8, =• |/(x) - /(j/)| < |

35i > 0 3 \/y 3 |x - y\ < S 2 => \g(x) - g(y)\ < |

Let 8 = min{<5i,£ 2 } > 0.

My 9 |x - 2/| < J =• |/(x) - /(j/)| < |, | 5 (x) - g(j/)| < 1

l(/ + </)(*) - (/ + </)(*) I = I/O*) + <K*) - f(v) - g(v)\ <

\f(x)-f(y)\ + \g(x)-g(y)\<^ + ^e

Thus, Ve > 0 3d > 0 9 Vy 9 |x - 2/| < 8 =» |(/ + <?)(x) - (/ + fl)(j/)| < e.

Therefore, / + g is continuous at x.

(b) / is continuous at x:

Vei > 0 3<5 > 0 9 My 3 |x - j/| < 8 =» |/(x) - /(j/)| < e.

Fix e = e. Then,

3J > 0 (say 8) 3 My (can fix at y) 9 |x — y\ < 8 =>• |/(x) — /(j/)| < e.

We have |x — f/| < 5 =>• |/(x) — /(j/)| < £•

Vy 9 |x - y\ < 8, |/(x) - f(y)\ <c\x-y\.

Choose y 3 \x — y\ < 8, \f(x) — f(y)\ < c\x — y\ < c6.

If <. c- >, we will reach the desired condition. One can choose 0 < 8 <

min {£,§}.

My 9 \x - y\ < 5 < 6, |/(x) - f(y)\ < c\x - y\ < cS < e.