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Limit x going to 3 (squareroot(x)-squareroot(3))/x-3
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as X -> 3 this equation begins to approach infinity. You can evaluate it at x = 3.000000001 or 2.9999999 to get an approximate answer.
So do this problem as if x was approaching infinity?
hi do you know the Lhospitals rule?
I do not
Limit x going to 3 (squareroot(x)-squareroot(3))/x-3 =Lim x->3 (squareroot(x)-squareroot(3))/x-3 =Lim x->3 ((1/2)x^-1/2) -0/(1) =Lim x->3 (1/(2x^1/2) =1/(2sqrt3)=approx 0.28867 ans
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