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

1 Answer

Dec 20, 2016

#lim_(x->2^+)(2-x)/(x^2-4)=-1/4#

Explanation:

Note that we can write the function as:

#(2-x)/(x^2-4)= - (x-2)/((x-2)(x+2))= -1/(x+2)#

In this way we eliminate the discontinuity for #x=2# and find that:

#lim_(x->2^+)(2-x)/(x^2-4)=lim_(x->2^+) -1/(x+2) =[-1/(x+2)]_(x=2)=-1/4#

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