I am trying to find where the graph of $f(x)=x^{2/3}-(x-1)^{1/3}$ has a vertical tangent line. I then need to confirm my findings using limit calculations. This is a problem from a book that I am doing for fun.
Plotting the graph shows what appears to be a vertical tangent line at $x=1$. The plot is seen in the following image:
Now I would like to confirm using limit calculations. I have
\begin{align*}
\lim_{h\rightarrow 0}\frac{f(x_0+h)-f(x_0)}{h}&=\lim_{h\rightarrow 0}\frac{\left[(1+h)^{2/3}-(1 + h – 1)^{1/3}\right]-\left[1^{2/3}-(1-1)^{1/3}\right]}{h}\\
&=\lim_{h\rightarrow 0}\frac{(1+h)^{2/3}-h^{1/3}-1}{h}.
\end{align*}
At this point, I'm stuck, as I don't know how to proceed. The term $(1 + h)^{2/3}$ in particular is giving me problems. Obviously, from the graph, I am expecting a result of $\pm\infty$ for the limit, but I just don't know what to do. Any help would be much appreciated.
Best Answer
Note that$$\lim_{h\to0}\frac{(1+h)^{2/3}-1}h\tag1$$is the derivative at $1$ of $x\mapsto(1+x)^{2/3}$ (which is $\frac23$). So, you only need to see that$$\lim_{h\to0}\frac{h^{1/3}}h=\lim_{h\to0}h^{-2/3}=\infty.$$
If you want to avoid differentiation, you can prove that $(1)$ is equal to $\frac23$ as follows:\begin{align}(1)&=\lim_{h\to0}\frac{\left((1+h)^{2/3}-1\right)\left((1+h)^{4/3}+(1+h)^{2/3}+1\right)}{h\left((1+h)^{4/3}+(1+h)^{2/3}+1\right)}\\&=\lim_{h\to0}\frac{(1+h)^2-1}{h\left((1+h)^{4/3}+(1+h)^{2/3}+1\right)}\\&=\lim_{h\to0}\frac{h+2}{(1+h)^{4/3}+(1+h)^{2/3}+1}\\&=\frac23.\end{align}