Limits problem using trigonometric identities

[shirkay]

Hello, I have been working on this problem for hours trying all kinds of different trig identities and cannot find the limit!

lim [(1-cosx)^2]/[tg^3(x)-sin^3(x)]
x->0

help will be much appreciated!

 

[mmm4444bot]

You certainly do not need to post hours' worth of efforts. Just show us one effort, as described in our posting guidelines. Cheers

Also, what is the meaning of tg^3(x) ? Is that supposed to be tan^3(x) ?

 

[shirkay]

tg is tan... for some reason here in Israel they use "tg".

These were my efforts thus far:

(1-cosx)^2 ------> (1-cosx)(1+cosx)-----> [2sin^2(x/2)](1+cosx)

for the denomenator I tried several things... changing tg to sin/cos,
or changing tan^3x-sin^3x -----> (tanx-sinx)(tan^2x+tanxsinx+sin^2x)
I tried bringing it to a point where I could use the (sinx)/x = 1 rule... didn't work either. I just couldnt bring anything to cancel out and kept getting a 0/0 limit.

 

[MarkFL]

I would write:

\(\displaystyle \lim_{x\to0}\dfrac{(1-\cos(x))^2}{\tan^3(x)-\sin^3(x)}=\lim_{x\to0}\dfrac{(1-\cos(x))^2}{\tan^3(x)(1-\cos^3(x))}=\)

\(\displaystyle \lim_{x\to0}\dfrac{(1-\cos(x))^2}{\tan^3(x)(1-\cos(x))(1+\cos(x)+\cos^2(x))}=\lim_{x\to0}\dfrac{1}{1+\cos(x)+\cos^2(x)}\cdot\lim_{x\to0}\dfrac{1-\cos(x)}{\tan^3(x)}=\)

\(\displaystyle \frac{1}{3}\lim_{x\to0}\dfrac{1-\cos(x)}{\tan^3(x)}\)

From here, application of L'Hôpital's rule twice will get you the result.

 

[shirkay]

I would write:

\(\displaystyle \lim_{x\to0}\dfrac{(1-\cos(x))^2}{\tan^3(x)-\sin^3(x)}=\lim_{x\to0}\dfrac{(1-\cos(x))^2}{\tan^3(x)(1-\cos^3(x))}=\)

\(\displaystyle \lim_{x\to0}\dfrac{(1-\cos(x))^2}{\tan^3(x)(1-\cos(x))(1+\cos(x)+\cos^2(x))}=\lim_{x\to0}\dfrac{1}{1+\cos(x)+\cos^2(x)}\cdot\lim_{x\to0}\dfrac{1-\cos(x)}{\tan^3(x)}=\)

\(\displaystyle \frac{1}{3}\lim_{x\to0}\dfrac{1-\cos(x)}{\tan^3(x)}\)

From here, application of L'Hôpital's rule twice will get you the result.

We haven't learned L'Hopital's rule yet...

 

[MarkFL]

We haven't learned L'Hopital's rule yet...

If I had a dollar for every time I've heard that...

Well, you can try the series expansion at \(\displaystyle x=0\).

 

[Deleted member 4993]

Hello, I have been working on this problem for hours trying all kinds of different trig identities and cannot find the limit!

lim [(1-cosx)^2]/[tg^3(x)-sin^3(x)]
x->0

help will be much appreciated!

I get limit DNE (Does not exist = m/0 form).

Would that be acceptable answer for your problem?

If not - you may want to check your original post mistakes.

 

[lookagain]

tg is tan... for some reason here in Israel they use "tg".

These were my efforts thus far:

(1-cosx)^2 ** equals sign (1-cosx)(1+cosx) <----- This is not true.

equals sign [2sin^2(x/2)](1+cosx) <----- So then the second and third expressions
would not be equal.

for the denomenator I tried several things... changing tg to sin/cos,
or changing tan^3x-sin^3x equals sign (tanx-sinx)(tan^2x+tanxsinx+sin^2x)


I tried bringing it to a point where I could use the lim (x -> 0) (sinx)/x = 1 rule... didn't work either. I just couldnt bring anything to cancel out and kept getting a 0/0 limit.

\(\displaystyle ** \ [1 - cos(x)]^2 \ = \ [1 - cos(x)][1 - cos(x)]\)

\(\displaystyle But \ \ 1 - cos^2(x) \ = \ [1 - cos(x)][1 + cos(x)]\)

 

[Deleted member 4993]

Hello, I have been working on this problem for hours trying all kinds of different trig identities and cannot find the limit!

lim [(1-cosx)^2]/[tg^3(x)-sin^3(x)]
x->0

help will be much appreciated!

[(1-cosx)^2]/[tg^3(x)-sin^3(x)]

= [(1-cosx)²]/[sin³(x){sec³x -1}]

= -[(1-cosx)]/[sin²(x)tan(x){sec²x + sec(x) + 1}]

= -[(2sin²(x/2)]/[4sin²(x/2)*cos²(x/2)*tan(x){sec²x + sec(x) + 1}]

= -[1]/[2*cos²(x/2)*tan(x){sec²x + sec(x) + 1}]