Compute limit $\lim\limits\frac{\sqrt[2017]{2x-1}-x^{2017}}{x-1}_{x\rightarrow1}$

Question

Compute limit $$\lim_{x\to 1}\frac{\sqrt[2017]{2x-1}-x^{2017}}{x-1}$$

Answer 1

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, $$\lim_{x\to 1}\frac{\sqrt[2017]{2x-1}-x^{2017}}{x-1}$$ $$=\lim_{x\to 1}\frac{\frac{2}{2017}(2x-1)^{-\frac{2016}{2017}}-2017\cdot x^{2016}}{1}$$ $$=\frac{\frac{2}{2017}(2-1)^{-\frac{2016}{2017}}-2017}{1}$$ $$=\frac{2-2017^2}{2017}$$

Answer 2

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?

Answer 3

Without L'Hospital: Substitute $u=x-1$: it's $$\lim_{u \to 0} \frac{(2u+1)^{1/2017} - (u+1)^{2017}}{u}$$

This is the derivative of $u \mapsto (2u+1)^{1/2017} - (u+1)^{2017}$ at $u=0$. You can find it using the chain rule.

Answer 4

$$\lim_{x\to 1}\frac{\sqrt[2017]{2x-1}-x^{2017}}{x-1}$$

$$=\lim_{x\to 1}\frac{\sqrt[2017]{2x-1}-1}{x-1}-\lim_{x\to 1}\dfrac{x^{2017}-1}{x-1}$$

For the first limit, set $\sqrt[2017]{2x-1}=v+1$

$$\lim_{x\to 1}\frac{\sqrt[2017]{2x-1}-1}{x-1}=2\lim_{v\to0}\dfrac v{\left((1+v)^{2017}-1\right)}=2\lim_{v\to0}\dfrac v{2017v+O(v^2)}=?$$

For the second limit set $x-1=u\iff x=1+u$

Or for $x-1\ne0,x^{2017}-1=x^{2016}+x^{2015}+\cdots+x+1$

Answer 5

To simplify typing we replace $2017$ by $n$ and then we have \begin{align} L &= \lim_{x \to 1}\frac{(2x - 1)^{1/n} - x^{n}}{x - 1}\notag\\ &= \lim_{x \to 1}\frac{(2x - 1)^{1/n} - 1}{x - 1} - \frac{x^{n} - 1}{x - 1}\notag\\ &= \lim_{t \to 1}2\cdot\frac{t^{1/n} - 1}{t - 1} - n\text{ (putting }2x - 1 = t)\notag\\ &= \frac{2}{n} - n\notag\\ &= \frac{2 - n^{2}}{n}\notag\\ &= -\frac{4068287}{2017}\notag \end{align} Note that we have used the quite unpopular (at least on MSE) standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$