Math question, please help :) (1 Viewer)

[Pote]

Can anyone please tell me how to do it? I took lim... but somehow I just couldn't get the answer :S

The blurry bit says at end of question a) lim(h->0) log (1 + h/x)^(1/h)

thank you so much.

 

[acmilan]

Let f(x) = lnx

f(x+h) = ln(x+h)

y'
= lim (h->0) (f(x+h) - f(x))/h
= lim (h->0) (ln(x+h)-lnx)/h
= lim (h->0) (ln[(x+h)/x])/h
= lim (h->0) (ln[1 + h/x])/h
= lim (h->0) (1/h)*(ln[1 + h/x])
= lim (h->0) (ln[1 + h/x])^1/h (the number in front of a log becomes the power)

 

[Pote]

ohhh thanks -_- damn, i feel stupid now.

for question b) ... how do i answer it in a correct format? do i just say y' = lim.... = 1/x

=S

 

[acmilan]

I dont know if they want you to do anything fancy, but if y' = 1/x and y' = that limit, then wouldnt it imply 1/x = that limit? I think thats all you need.

 

[Pote]

thanks!

 

[justchillin]

I can assure you, however, that no first principals will be examined in 2 unit yet alone 3 unit...

 

[SaHbEeWaH]

I can assure you, however, that no first principals will be examined in 2 unit yet alone 3 unit...

first principals is extension 1 prelim work and so it can be (and has been) tested in hsc
2u never learn it, they just use the formulas

 

[acmilan]

Even so, its great practice for first year uni maths, where first principles are your friend, especially in double variable calculus