How do you find the limit of #(x^2-4)/(x-2)# as x approaches 2?

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Answer 1 · 2016-11-03T21:44:43

# lim_(xrarr2)(x^2-4)/(x-2) = 4 #

If we look at the graph of #y=(x^2-4)/(x-2) # we can see that it is clear that the limit exists, and is approximately #4#

graph{(x^2-4)/(x-2) [-10, 10, -5, 5]}

The numerator is the difference of two squares, and as such we can factorise using it as

# A^2-B^2 -= (A+B)(A-B) #

Se we can factorise as follows:

# lim_(xrarr2)(x^2-4)/(x-2) = lim_(xrarr2)(x^2-2^2)/(x-2) #

# = lim_(xrarr2)((x+2)(x-2))/(x-2) #
# = lim_(xrarr2)((x+2)cancel(x-2))/cancel(x-2) #
# = lim_(xrarr2)(x+2) #
# = (2+2) #
# = 4 #

Which is completely consistent with the above graph.