find the slope of the tangent line Sorry I couldn't figure out how to use the editor right now and I really need help with this question. 
Let $f(x)=\frac{1}{\sqrt[3]{x}}$. find the slope of the tangent line at $(8,\frac{1}{2})$
So far this is what I have I just don't know what to do from here
$$
\begin{array}{lll}
m&=&\lim_{h \to 0}\frac{f(8+h)-f(8)}{h}\\
&=&\lim_{h \to 0} \frac{\frac{1}{\sqrt[3]{8+h}}-1/2}h \\
&=&\lim_{h \to 0} \ 2-\frac{\sqrt[3]{8+h}}{2\sqrt[3]{8+h}}\\
&=&\lim_{h \to 0} \frac{2-\sqrt[3]{8+h}}{2h\sqrt[3]{8+h}}\\
\end{array}
$$
Can someone please help me out please 

From where you're at, do this trick:
\begin{align*}
&\lim_{h \to 0} \frac{2-\sqrt[3]{8+h}}{2h\sqrt[3]{8+h}}=\lim_{h \to 0} \frac{\left(2-\sqrt[3]{8+h}\right)}{2h\sqrt[3]{8+h}}\frac{\left((8+h)^{2/3} +2(8+h)^{1/3} + 4)\right)}{\left((8+h)^{2/3} +2(8+h)^{1/3} + 4)\right)}=\\
&\lim_{h\to 0}\frac{-h}{2h(8+h)^{1/3}\left((8+h)^{2/3} +2(8+h)^{1/3} + 4)\right)}= \\
&\lim_{h\to 0} -\frac{1}{2(8)^{1/3}(8^{2/3}+2\cdot8^{1/3}+4)}=-\frac{1}{48}
\end{align*}
Here's an equivalent limit proof.
\begin{align*}
&f'(8)=\lim_{x\to 8}\frac{f(x)-f(8)}{x-8} = \lim_{x\to 8}\frac{\frac{1}{x^{1/3}}-\frac{1}{2}}{(x^{1/3}-2)(x^{2/3}+2x^{\frac{1}{3}}+4)}= \\
&\lim_{x\to 8} \frac{\frac{2-x^{1/3}}{2x^{1/3}}}{(x^{1/3}-2)(x^{2/3}+2x^{1/3}+4)}=-\frac{1}{2}\lim_{x\to 8}\frac{1}{x^{1/3}(x^{2/3}+2x^{1/3}+4)}=\\
&-\frac{1}{2}\left(\frac{1}{2(4+4+4)}\right)=-\frac{1}{48}
\end{align*}

