Compute limit $\lim\limits\frac{\sqrt[2017]{2x-1}-x^{2017}}{x-1}_{x\rightarrow1}$ Compute limit $$\lim_{x\to 1}\frac{\sqrt[2017]{2x-1}-x^{2017}}{x-1}$$
L'Hospital's rule works fine here
$$
\lim_{x\to 1}\frac{\sqrt[2017]{2x-1}-x^{2017}}{x-1}=\lim_{x\to 1}\frac{\frac2{2017}\cdot(2x-1)^{-\frac{2016}{2017}}-2017x^{2016}}{1}.
$$
Can you finish it?
The limit $\lim_{x\to 1}\frac{\sqrt[2017]{2x-1}-x^{2017}}{x-1}$ is of the form $\frac00$ which is indeterminate. Hence, we can apply L'Hospital's Rule here.

Using L'Hospital's Rule,
$$\lim_{x\to 1}\frac{\sqrt[2017]{2x-1}-x^{2017}}{x-1}$$
$$=\lim_{x\to 1}\frac{\frac{2}{2017}(2x-1)^{-\frac{2016}{2017}}-2017\cdot x^{2016}}{1}$$
$$=\frac{\frac{2}{2017}(2-1)^{-\frac{2016}{2017}}-2017}{1}$$
$$=\frac{2-2017^2}{2017}$$