Prove 2x^2+5 is continuous at x=3

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Homework Statement



Using the de finition of continuity show that f(x) = 2x2 + 5 is continuous at x = 3

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The Attempt at a Solution


For all ε>0 there exists δ>0 such that |x-3|<δ implies that |2x2 + 5 -23| = |2(x2-9)| = |2(x+3)(x-3)| < |2(x+3)δ|

Could anybody give me advice as to where I should take it from here to show that |2x2 + 5 -23|<ε
 
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hi gottfried! :smile:

if eg you insist that δ < 1, then |x+3| is between 5 and 7 :wink:
 
|2(x+3)(x-3)| = 2|x+3||x-3| < 2δ|x+3|

Now is |x+3| = |x - 3 + 6| ? Also remember to use the triangle inequality.
 
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