Evaluate the following limits
Lim as x ->0 (2 to the power of x+3) subtract 8 divided by (2 to the power of x) subtract 1
lim as x ->0 (x + 8 to the power of 1/3) - 2 divided by x
Any help?
$\displaystyle \lim_{x \to 0} \frac{{(x + 8)^{1/3} - 2}}
{x}.$
Substitute $\displaystyle u^3=x+8,$
$\displaystyle \lim_{x \to 0} \frac{{(x + 8)^{1/3} - 2}}
{x} = \lim_{u \to 2} \frac{{u - 2}}
{{u^3 - 8}} = \lim_{u \to 2} \frac{1}
{{u^2 + 2u + 4}}.$
The rest follows.
Yes