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$$\lim_{x\to1}\frac{\sqrt{2-x}-1}{1+\sqrt[5]{x-2}}$$

I'm solving this without L'Hopital.

The first thing in my mind is to use variable substitution, that is, create some

$$w^{10} = 2-x$$

And then create a new limit where $w\to1$.

However, that's not possible because the radicands in both roots are different... Yet, they look so similar. I guess I cannot take out a $-1$ as common factor from one of them, because then I would have $\sqrt{-1}$...

So I guess it's not possible via variable substitution... or is it?


The next likely approach would be rationalization via conjugate. That is,

$$\frac{\sqrt{2-x}-1}{1+\sqrt[5]{x-2}}\cdot\frac{\sqrt{2-x}+1}{\sqrt{2-x}+1}$$

Which results in

$$\frac{2-x-1}{(1+\sqrt[5]{x-2})\cdot(\sqrt{2-x}+1)}$$

But that will still evaluate to $\frac{0}{0}$ and I don't see a way to further simplify this.

I would have tried to rationalize using the conjugate of the denominator, but I have a feeling that I'm not supposed to be doing this with fifth roots.


What should be my approach without L'Hopital? The answer has to be

$$\frac{-5}{2}$$

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HINT:

As lcm$(2,5)=10,$

let $\displaystyle\sqrt[10]{2-x}=u$ and $u\to1>0$

$\implies2-x=u^{10}$

$\displaystyle\implies\sqrt{2-x}=u^5$

and $\displaystyle x-2=-u^{10}\implies\sqrt[5]{x-2}=-u^2$

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  • $\begingroup$ Here's my attempt:$$w^{10} = 2-x \text{ therefore } w \to 1$$ then $$\lim_{w\to1}\frac{\sqrt{w^{10}} - 1}{1 + \sqrt[5]{-w^{10}}} = \frac{w^5-1}{1-w^2}$$ That is factorized to $$\frac{(w-1)(w^4+w^3+w^2+w)}{(w-1)(-w-1)}$$However $$\frac{1+1+1+1}{-1 - 1} = -2$$ $\endgroup$ – Zol Tun Kul Oct 1 '15 at 6:50
  • $\begingroup$ Ahh... My numerator factorization was wrong. It had to be $(w-1)(w^4+w^3+w^2+w+1)$, so at end it would indeed be $\frac{-5}{2}$. Thanks. $\endgroup$ – Zol Tun Kul Oct 1 '15 at 7:18
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Use of standard limits in case of evaluation of limits of algebraic expressions is not very popular, but it is the simplest approach to handle such problems.

We will use the following limit $$\lim_{x\to a} \frac{x^{n} - a^{n}} {x-a} =na^{n-1}\tag{1}$$ We have \begin{align} L&=\lim_{x\to 1}\frac{\sqrt{2-x}-1}{1+\sqrt[5]{x-2}}\notag\\ &= \lim_{t\to 1}\frac{t^{1/2}-1}{1-t^{1/5}}\text{ (putting }t=2-x) \notag\\ &= - \lim_{t\to 1}\frac{t^{1/2}-1}{t-1}\cdot\frac{t-1}{t^{1/5}-1}\notag\\ &= - \frac{1/2}{1/5}=-\frac{5}{2}\text{ (using (1))}\notag \end{align}

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$$\begin{aligned}\lim _{t\to 0}\left(\frac{\sqrt{2-\left(t+1\right)}-1}{1+\sqrt[5]{\left(t+1\right)-2}}\right) \\& = \lim _{t\to 0}\left(\frac{\sqrt{-t+1}-1}{1+\sqrt[5]{t-1}}\right) \\& = \lim _{t\to 0}\left(\frac{1-\frac{1}{2}t+o\left(t\right)-1}{1-1+\frac{1}{5}t+o\left(t\right)}\right) \\& = \lim _{t\to 0}\left(\frac{-\frac{1}{2}t+o\left(t\right)}{\frac{1}{5}t+o\left(t\right)}\right) \\& = \color{red}{-\frac{5}{2}} \end{aligned}$$ Solved with Taylor expansion

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