Asked by Henry
Find the limit when x approaches 0 3x^2+5/x+1
Answers
Answered by
oobleck
(3x^2 + 5)/(x+1)
the function is continuous everywhere except x = -1
since the denominator is not zero when x=0, just plug in x=0.
The limit is 5/1 = 5
the function is continuous everywhere except x = -1
since the denominator is not zero when x=0, just plug in x=0.
The limit is 5/1 = 5
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