Question

evaluate this limit

Answer 1

\[\huge \lim{_{x \rightarrow 1}\frac{\sqrt[5]{x}-1}{\sqrt{x}-1}}\]

Answer 2

Where is what I attempted. \[\large u = \sqrt[5]{x} -1 \]
so then \[(u-1)^{5} = x \]

Answer 3

Now I am stuck. How do I rewrite the bottom in terms of u and how do I rewrite the limit in terms of u?

Answer 4

Anybody Have any ideas?

Answer 5

do you know (or can you use) this :
\(\Large \lim \limits_{x \rightarrow a}\dfrac{x^n-a^n}{x-a}=n \times a^{n-1}\)
??

Answer 6

What is that rule?

Answer 7

i don't think it has a name...

Answer 8

So, I let \[f(x) = \sqrt{x}\]\[so, f(1) = 1\]

Answer 9

if you can't use that, can you use L'Hopital's rule ?

Answer 10

I'm not allowed to use that for the exam that is coming up. We have not learned that in class yet.

Answer 11

Is it like taking the derivate like \[\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a}\]

Answer 12

\[(x-1)=(\sqrt{x}-1)(\sqrt{x}+1)\]

\[(x-1)=(\sqrt[5]{x}-1)((\sqrt[5]{x})^4+(\sqrt[5]{x})^3+(\sqrt[5]{x})^2+\sqrt[5]{x}+1)\]

Answer 13

\[\frac{\sqrt[5]{x}-1}{\sqrt{x}-1}=\frac{(x-1)/((\sqrt[5]{x})^4+(\sqrt[5]{x})^3+(\sqrt[5]{x})^2+\sqrt[5]{x}+1)}{(x-1)/(\sqrt{x}+1)}\]

Answer 14

\[=\frac{\sqrt{x}+1}{(\sqrt[5]{x})^4+(\sqrt[5]{x})^3+(\sqrt[5]{x})^2+\sqrt[5]{x}+1}\]

as \(x\to 1\) you get

\[\frac{1+1}{1+1+1+1+1}=\frac{2}{5}\]

Answer 15

What just happened. I think we can use substitution.

Answer 16

\(x^n-a^n =(x-a)(x^{n-1}+ax^{n-2}+...+a^{n-1})\)
you used this, right ?

Answer 17

let me just take this in. you used the factoring method. and than evaluated the limit.

Answer 18

yes

Answer 19

I didn't know you could write It like that. I don't think I will remember that method in an exam.

Answer 20

replace a with 1 and \(x\) with \(\sqrt[n]{x}\)

Answer 21

1/5 -1 = -4/5 , (not 4/5)
1/5-2= -3/4, (not 3/4)
?

Answer 22

alright lets try that .\[Let f(x) = \sqrt{x} = u \]
\[f(1) = 1 , so, a = 1 \] now we can write \[\lim_{u \rightarrow 1}\frac{u^{\frac{1}{5}} -a}{u-a}\]

Answer 23

you could also let \[u=\sqrt[10]{x}\]

then \[u^{10}=x\]

Answer 24

then expand

Answer 25

so we know that \[\frac{d}{du}= \frac{1}{5}u * \frac{du}{dx} \]

Answer 26

mebs, if sqrt x = u
then x^1/5 will not = u^1/5 ...

Answer 27

\[\lim_{x \to1}\frac{\sqrt[5]{x}-1}{\sqrt{x}-1}=\lim_{u\to 1}\frac{u^2-1}{u^5-1}=\lim_{u\to 1}\frac{(u-1)(u+1)}{(u-1)(u^4+u^3+u^2+u+1)}\]

\[=\lim_{u\to 1}\frac{u+1}{u^4+u^3+u^2+u+1}=\frac{2}{5}\]

Answer 28

thought it is really just the same idea i did above

Answer 29

*though

Answer 30

well guys... I need time to interpret this. Thank a whole lot for all these ideas. I can't give you all a medal but thanks anyhow. @Zarkon and @hartnn

Answer 31

just remember that we used this here :
\(x^n-a^n =(x-a)(x^{n-1}+ax^{n-2}+...+a^{n-2}x+a^{n-1})\)

Answer 32

Noted*

Answer 33

how did Zarkon make the \[\sqrt[5]{x} = u^{2}\] @hartnn

Answer 34

by plugging in x=u^10

x^1/10 =u
squaring both sides
...

Answer 35

5th root of x = x^1/5

Answer 36

\[\sqrt[5]{x}=\sqrt[5]{u^{10}}=(u^{10})^{1/5}=u^{(10\cdot\frac{1}{5})}=u^2\]

Answer 37

O I see, I just have to use the exponent rule of \[(x^{n})^{a} = x^{n*a}\]

Answer 38

YES I GET IT ! I am going to attempt a similar problem and than I'll post it to see if you guys get what I get. @Zarkon and @hartnn

Answer 39

sure :)

Answer 40

just for another method that i proposed :
\(\Large \lim \limits_{x \rightarrow a}\dfrac{x^n-a^n}{x-a}=n \times a^{n-1}\)

\(\huge \lim \limits_{x \rightarrow 1}{\frac{\sqrt[5]{x}-1}{\sqrt{x}-1}}=\huge \lim \limits_{x \rightarrow 1}\dfrac{\frac{\sqrt[5]{x}-1}{{x}-1}}{\frac{\sqrt x-1}{x-1}}=\dfrac{1/5}{1/2}=2/5\)