Question

\[\lim_{x \rightarrow 0}\left(\begin{matrix}(1+x)^{7}-(1-x)^{7} \\ (1+x)^{5}-(1-x)^{5}\end{matrix}\right)\]

Answer 1

a good guess will solve it immediately

Answer 2

@satellite73 : lol and i need that guess! =.=

Answer 3

is this supposed to be a fraction and not a matrix?

Answer 4

\[\lim_{x\to0}{(1+x)^7-(1-x)^7\over(1+x)^5-(1-x)^5}\]?

Answer 5

@TuringTest :yes it should be a fraction but i dont know how to make a fraction

Answer 6

@TuringTest :yup thats it !

Answer 7

(1^7-1^7)/(1^5-1^5) Not sure you're able to do it like this, just guessing! :D

Answer 8

@Noliec no that's not how you do it, and that gives 0/0
please don't guess like that

Answer 9

I guess you would have to expand all factors and solve it that way?

Answer 10

no there is a better way
I unfortunately feel like the answer is escaping me... embarrassing :(

Answer 11

something to do with difference of squares I'm sure

Answer 12

how about L'hopital rule..

Answer 13

@TuringTest :I think that w should distribute the power ,like make the 7th power =(1-x)^2 (1-x)5

Answer 14

7/5

Answer 15

L'hospital will get very messy I think
@Eyad but then what...?
wolfram has its answer by expanding, so that is one way
but there must be a better one

Answer 16

yes that's the answer @cinar but how did you get it without the wolf?

Answer 17

nope.. L'hopital

Answer 18

ok let me try it
I thought it would be messy....

Answer 19

I still get 0/0 with l'hospital
I must be missing something

Answer 20

\[\lim_{x \rightarrow 0} \frac{7(1+x)^6+7(1-x)^6}{5(1+x)^4+5(1-x)^4}=7/5\]

Answer 21

how did the negative turn positive o-0 ?

Answer 22

oh nevermind

Answer 23

because \[y= f(x)^n\]
\[y'=n f(x)^{n-1}*f'(x)\]

Answer 24

duh
ok

Answer 25

funny how wolf didn't use l'hospital
usually it uses it every chance it gets
oh well, thanks cinar!

Answer 26

Thanks @cinar

Answer 27

np