What does the l equals to?

Question

I have got this limit and I have no clue how to resolve it:

$$l=\lim_{x\to 0}\frac{ \sqrt[2018]{1+x}-\sqrt[2019]{1-x}}x $$

Can you give me any tips or even resolve it please?

Answer 1

Hint: Compute $$\lim_{x\to 0}\frac{\sqrt[2018]{1+x}-\sqrt[2018]{1+0}}{x}$$ and $$\lim_{x\to 0}\frac{\sqrt[2019]{1-x}-\sqrt[2019]{1-0}}{x}$$

Answer 2

Hint: If you rewrite your expression to $$ \frac{\sqrt[2018]{1+x}-1}x - \frac{\sqrt[2019]{1-x}-1}x $$ the limit decomposes into the difference of two derivatives

Answer 3

If you know how to compute the derivative of $\sqrt[2018]{1+x}$ then de L'hôpital's rule gives you a straightforward answer.

Answer 4

use L'H,

$\;\;\;\;\lim_{x \to 0} \frac{\sqrt[2018]{1+x}-\sqrt[2018]{1-x}}{x}$

$=\lim_{x \to 0} \frac{\frac{1}{2018(1+x)^{\frac{2017}{2018}}}+\frac{1}{2018(1-x)^{\frac{2017}{2018}}}}{1}$

$=\frac{1}{2018}+\frac{1}{2018}$

$=\color{blue}{\frac{1}{1009}}$

Answer 5

$n:=2018$, then $n+1=2019$.

$f(x) =\dfrac{(1+x)^{1/n}-1}{x} -$

$\dfrac{(1-x)^{1/(n+1)} -1}{x}$.

1) $\lim_{x \rightarrow \infty} \dfrac{(1+x)^{1/n} -1}{x}=(1/n)$

2) $\lim_{x \rightarrow \infty} \dfrac{(1-x)^{1/(n+1)}-1}{x}=$

$-1/(n+1).$

Note : The limits are the derivatives of the resp. functions at $x=0$.

Combining:

$ \lim_{x \rightarrow 0} f(x)= 1/n+1/(n+1)$.