Limit problems

[steve_254]

ok I have no idea where to start or whats going on in the following 2 limits

limx->0 [ 2tanX / sqr(1+X) - 1 ]

and

limx->0 [ X^2(sqr(1+x)) / 1 - cos3x ]

any help at all would be greatly appreciated

 

[Guest]

limx->0 [ 2tanX / sqr(1+X) - 1 ]

let u = 2tanx

v= sqr(1+x)-1


dy/dx = (v(du/dx)- u (dv/dx)) / v^2

this is the method have a go.

 

[steve_254]

bleh i dont know what that means haha
owell

 

[stapel]

Which part didn't you understand? Please be specific in stating what you've done and where you're stuck. Thank you.

Eliz.

 

[soroban]

Hello, steve_254!

I'll walk through #2 . . .

Are you familiar with this limit theorem?

. . . . . . sin θ
. . . lim .------ . = . 1
. . θ->0 . .θ

2) lim(x->0) [ x^2 sqr(1+x) ] / [1 - cos3x ]

. . . . . . . . . . . . ._____
. . . . . . . . . . x²√1 + x
We have: . --------------
. . . . . . . . . . 1 - cos3x


Multiply top and bottom by (1 + cos3x):
. . . . . . ____ . . . . . . . . . . . . . . . . . . . ____ . . . . . . . . . . . . . . . .____
. . . x²√1 + x . . .1 + cos3x . . . . . . x²√1 + x (1 + cos3x) . . . .x²√1 + x (1 + cos3x)
. . -------------- . -------------- . = . -------------------------- . = . --------------------------
. . .1 - cos3x . . .1 + cos3x . . . . . . . . 1 - cos²3x . . . . . . . . . . . . . sin²3x

. . . . . . . . . . .x² . . . ____
We have: . -------- √1 + x (1 + cos3x)
. . . . . . . . . sin²3x

. . . . . . . . . . . . . .1 . . .9x² . . .____
Multiply by 9/9: . -- . --------- √1 + x (1 + cos3x)
. . . . . . . . . . . . . .9 . .sin²3x


. . . . . . . . . 1 .|. .3x . |². ____
We have: . -- | ------- | √1 + x (1 + cos3x)
. . . . . . . . . 9 .| sin3x |


Now take the limit as x -> 0 (and 3x -> 0):
. . . . . . . . . . . . . . .1 . . . . ____ . . . . . . . . . . . . . 1 . . . . . . . . . . . . 2
. . . and we get: . -- (1²)√1 + 0 (1 + cos 0) . = . -- (1) (1) (2) . = . --
. . . . . . . . . . . . . . .9 . . . . . . . . . . . . . . . . . . . . . .9 . . . . . . . . . . . . 9

[Someone check my work . . . please!]

 

[steve_254]

sweet thanks guys.... i figure the first one out too thanks fr the help