lim (x --> -8) of:
(square root(1-x)) - 3 divided by 2 + (cube root(x))
I believe I should multiply by the congugate, but I'm not getting anywhere.
Thanks for any help
I suggest that you first make the substitution $\displaystyle x = t^3$:
$\displaystyle \lim_{t \rightarrow -2} \frac{\sqrt{1 - t^3} - 3}{2 + t}$.
Now multiply numerator and denominator by the conjugate surd, factorise the numerator, divide numerator and denominator by the common factor and then take the limit.