In formally proving that lim x → − 5 ( 1/6 x + 9 ) = 49/6 , let ε > 0 be arbitrary. Determine δ as a function of

Question

Note: in this case  will be a function of . You will need to write the word epsilon for in the answerbox

Answer 1

Notice that

| ( 1/6 x + 9 ) - 49/6 | < ε

| ( 1/6 x + 5/6 | < ε

| ( x - (-5) | < 6ε

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Let δ = 6ε

If 0<| ( x - (-5) |<δ

| ( x - (-5) |<6ε

| ( 1/6 x + 5/6 | < ε

then | ( 1/6 x + 9 ) - 49/6 | < ε

For all ε>0, there exists δ>0 (δ = 6ε) such that if | ( x - (-5) |<δ then | ( 1/6 x + 9 ) - 49/6 | < ε.

We prove that lim x → − 5 ( 1/6 x + 9 ) exists and its value is 49/6 by taking δ = 6ε.