Question

Evaluate
\[\lim_{x \rightarrow 0} \frac{(1+x)^{\frac{1}{k}} -1}{x}\]

Answer 1

L'Hopital's rule not allowed ?

Answer 2

Yes, not allowed!

Answer 3

put 1+x = y

Answer 4

then use
\[\lim_{y \rightarrow a}\frac{ y^n-a^n }{ y-a }=ny^{n-1}\]

Answer 5

Can you explain how you get that?

Answer 6

the formula ? its standard, like sin x/x=1
after substittution, u get lim y->1 y^(1/k)-1/(y-1)

Answer 7

which is just (1/k) y ^{1/k-1}
which you would expect by L'hopitals....

Answer 8

If you don't mind, I would like to know how you get that formula, since I haven't learnt that before, if I remember correctly.

Answer 9

sure, u know the general formula for a^n-b^n ??

Answer 10

\(a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+....b^{n-1})\)
use this :
y-a will cancel out
pt y=a in remaining polynomial.

Answer 11

tell me if u couldn't get it.

Answer 12

Yes!! I was trying to use that formula but seemed that I couldn't continue..
Since a = 1+x there, b =1 and n= 1/k
When I substitute a = 1+x, it seems that I have to ''expand'' (1+x)^(1/k -1). That would be a mess!

Answer 13

oh, yes. that can also be done.....
but y^n-a^n/y-a is quite standard.....

Answer 14

But it comes to a problem that I haven't learnt it.. And it's hard for me to write that if I don't understand that.. I'm sorry!!!

Answer 15

thats actually good spirit, keep it up.

Answer 16

Haha! Would you mind explaining the standard form??

Answer 17

but u need to know this relation:
\(a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+....b^{n-1})\)

Answer 18

Yes, I know it!

Answer 19

a-b cancelled out right ?

Answer 20

1+x-1 = x

Answer 21

Sure!!

Answer 22

ok,
so now directly put x=0, what u get ?

Answer 23

i am sorry, the formula is
\(\lim_{y \rightarrow a}\frac{ y^n-a^n }{ y-a }=na^{n-1}\)

Answer 24

1+1+1+...+1 ....

Answer 25

1/k times, right ?

Answer 26

Does that matter?

Answer 27

I suppose 1^(1/k) = 1 ..

Answer 28

For k =/= 0

Answer 29

yup, u need to know how many times '1' is added.

Answer 30

1^{1/k-1} + 1^{1/k-2}+....1^{1}

Answer 31

Ahh.. Yes.... 1/k times..

Answer 32

then it would be 1/k, isn't it ?

Answer 33

It is!

Answer 34

sorry for my typing mistake earlier....its na^{n-1} in that formula....
any more doubts here ?

Answer 35

I don't understand...

Answer 36

u got 1/k, its correct answer, which part did u not understand ?

Answer 37

na^{n-1} ...

Answer 38

that general formula, if u didn't understand, leave it...you were able to solve without that also, isn't it ?

Answer 39

Oh, you're right! Thanks!!~

Answer 40

welcome ^_^