limit
x tends to 9 x^3/2 - 27/x-9 ?
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Let √x be t now the limit function can be written as (t3 – 27)/(t2 – 9) In the above expression, t-3 is the common term cancelling it we get (t2 + 3t + 9)/(t+3) when x tends to 9 t tends to 3 so limit is (9 + 9 + 9)/(3+3) = 27/6 =4.5
thanks..
lim ( x^n-a^n)/x-a=na^(n-1)
x -> a
lim (x^3/2-27)/x-9 = 3/2*9^(3/2-1) = 4.5
x->9
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