How do you find the limit of #(sqrtx-3)/(x-9)# as #x->9#?

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Answer 1 · 2017-11-30T02:21:47

#lim_(x->9)(sqrtx-3)/(x-9)=1/6#

Note: We can't evaluate the limit in the form that it is initially. Doing so will make the fraction undefined.

We can rationalize the number by multiplying both the numerator and denominator by the numerator's conjugate.

#(sqrtx-3)/(x-9)*(sqrtx+3)/(sqrtx+3)#

#=(x+3sqrtx-3sqrtx-9)/((x-9)(sqrtx+3))#

#=cancel(x-9)/(cancel(x-9)(sqrtx+3))#

#1/(sqrtx+3)#

Now we can calculate the limit:

#lim_(x->9)1/(sqrtx+3)=1/(sqrt9+3)=1/(3+3)=1/6#