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does the limit as x approaches 3 of |x-3|/x-3 exist? If so, can you show how you came to that conclusion.
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no it doesn't
this is identically 1 if x > 3 and -1 if x < 3. limit from the left is -1, limit from the right is 1 and since those numbers are not the same, there is no limit at 3
thanks! i initially though that the limit didnt exist because i remember a previous example in class like this but i did not know how to prove it..thanks!
could you explain why it =1 if x>3 and why it is -1 if x<3 ?
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