Q1:

lim 2x^3 squareroot(x+6)

x->3

I substitute in the 3

=2(3)^3 squareroot(3+6)

=(54)(3)

=162

I'm not sure if 162 is the limit or not.. the number seems huge

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Q2:

lim (64-x^2)/(squareroot(8-x))

x-> negative infinity

First i found the greatest degree in the denominator

= squareroot of x

=((64/squareroot x)-(x^2/ squareroot x))/(squareroot (x(8/x-1))

=substitute in negative infinity

=-1/squareroot x

=0

Thanks