How do you find the limit of #sqrt(x-4)/(x-16)# as x approaches 16?

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Answer 1 · 2016-03-23T02:45:49

#1/8#

We can factor #x-16# as a difference of squares.

Since differences of squares take the pattern #a^2-b^2=(a+b)(a-b)#, we see that in #x-16#,

#a^2=x" "=>" "a=sqrtx#
#b^2=16" "=>" "b=4#

Yielding:

#x-16=(sqrtx+4)(sqrtx-4)#

The problem then becomes:

#lim_(xrarr16)(sqrtx-4)/(x-16)=lim_(xrarr16)(sqrtx-4)/((sqrtx+4)(sqrtx-4))=lim_(xrarr16)1/(sqrtx+4)#

Evaluate the limit by plugging in #16#:

#lim_(xrarr16)1/(sqrtx+4)=1/(sqrt16+4)=1/(4+4)=1/8#