Question 637748: Find limit ((x+1)^2-1)/ x
x->0
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Simplify the numerator
[(x+1)^2-1] = x^2 + 2x +1 -1 = x^2 + 2x = [x(x+2)]
Rewrite the given expression
[(x+1)^2-1]/x = [x(x+2)]/x = [x+2]
Limit as x goes to 0 of [x+2] = 0+2 = 2
Because the numerator had a factor x, it cancelled with x in the denominator prior to x reaching a limit of 0. Thus we never did divide by zero and the answer is a finite value of 2.
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