limit f(x) = L as x-->a
For every E>0 there exists Delta_E such that
L-E < f(x) < L+E when a-Delta_E < x < a+Delta_E
Working backwards:
If
16-E < 10-2x < 16+E
Then
"6-E < -2x < 6+E\\\\\n\n\n\n (1\/2)E - 3 > x > (-1\/2)E - 3\\\\\n\n\n\n -3 + (1\/2)E > x > -3 - (1\/2)E\\\\\n\n\n\n -3 - (1\/2)E < x < -3 + (1\/2)E\\\\\n\n\n\n -3 - E\/2 < x < -3 + E\/2\\\\"
For E>0, Let Delta_E = E/2
When -3-Delta_E < x < -3+Delta_E
"-3-E\/2 < x < -3 + E\/2\\\\\n\n \n\n -6 - E < 2x < -6 + E\\\\\n\n\n\n E+6 > -2x > 6-E\\\\\n\n\n\n E+16 > 10-2x > 16-E\\\\\n\n\n\n E+16 > f(x) > 16-E\\\\"
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