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

Question

7184 views around the world

You can reuse this answer
Creative Commons License

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#