How do you evaluate the limit of #lim ((x+1)^2-1)/(x+2)# as #x->0#?

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Answer 1 · 2016-12-13T02:17:01

Please see the explanation.

#lim_(xto0) ((x + 1)^2 - 1)/(x + 2) = ?#

Expand the square:

#lim_(xto0) (x^2 + 2x + 1 - 1)/(x + 2)#

Combine the constants in the numerator:

#lim_(xto0) (x^2 + 2x)/(x + 2)#

#x+2# is a common factor in the numerator and the denominator:

#lim_(xto0) x = 0#

#:.#

#lim_(xto0) ((x + 1)^2 - 1)/(x + 2) = 0#

Also, we could have evaluated the original expression at 0.