Limit of a trigonometric function

[chwala]

Homework Statement

Mod note: Edited the following to fix the LaTeX[/B]
compute

##\lim_{n \rightarrow +0} \frac {8-9cos x+cos 3x} {sin^4(2x)}####\lim_{n \rightarrow +\infty} \frac {\sin(x)} x##

##\lim_{n \rightarrow +\infty} \frac {\sin(x)} x##ok find limit as x→0 for the function ##[ 8-9cos x + cos 3x/{sin^4}2x]##

Homework Equations

The Attempt at a Solution

use Lhopital rule...

 

[BvU]

Homework Statement

compute

$$\lim_{n \downarrow 0} {\frac {8-9cos x+cos 3x} {sin^4}2x}$$$$\lim_{n \rightarrow +\infty} {\frac {\sin(x)} x}$$

$$\lim_{n \rightarrow +\infty} {\frac {\sin(x)} x}$$[/B]ok find limit as x→0 for the function ##f(x) = 8-9cos x + cos 3x/{sin^4}(2x) ##

Homework Equations

The Attempt at a Solution

use Lhopital rule...

Well, 2 and 3 look alike and don't need L'H rule. You can deal with those yourself, right ? If not, what's the problem ?
1 and 4 are one and the same too ?

 

[chwala]

Well, 2 and 3 look alike and don't need L'H rule. You can deal with those yourself, right ? If not, what's the problem ?
1 and 4 are one and the same too ?

sorry my interest is on 1 only...i was trying to post in latex form by using 2 and 3 as a hint...

 

[BvU]

No need to apologize, we just want to be clear on what we are looking at. So $$
\lim_{n \downarrow 0} {\frac {8-9cos x+cos 3x} {sin^4}(2x)} $$ is the right one ?

(as you see, it's easy to come to the wrong expression...)

 

[chwala]

ok find limit as x→0 for the function ##[ 8-9cos x + cos 3x/{sin^4}2x]##
note that 2x is on the denominator...wish i could post this using latex...

 

[BvU]

Now we are at $$
\lim_{n \downarrow 0} 8-9\; cos x+ { cos 3x \over sin^4 (2x)}
$$ which probably is not what your exercise intended.

 

[chwala]

not clear post in latex please

 

[BvU]

Now we are at $$\lim_{n \downarrow 0} 8-9\; cos x+ { cos 3x \over sin^4 (2x)} $$ which probably is not what your exercise intended.

I though I did post using LaTex ? What is it that is not clear ?

 

[chwala]

##sin^42x## is the common denominator to all those terms in numerator and not the way you have posted. It is post number 4 which is correct with the only amendment on the 2x, that should be on the denominator as ## sin^4.2x##

 

[member 587159]

##sin^42x## is the common denominator to all those terms in numerator and not the way you have posted. It is post number 4 which is correct with the only amendment on the 2x, that should be on the denominator as ## sin^4.2x##

Well you have posted it without the nessecary brackets so he couldn't know. He just copies what you wrote in the first post.

 

[BvU]

Good. So we are at $$
\lim_{n \downarrow 0} \frac {8-9cos x+cos 3x} {sin^4(2x)}
$$
(the 2x in the wrong place was caused by me unintentionally ).

You mentioned l'Hopital. It is so to say your attempt at solution. What is it and what does it yield here ?

 

[member 587159]

Good. So we are at $$
\lim_{n \downarrow 0} \frac {8-9cos x+cos 3x} {sin^4(2x)}
$$
(the 2x in the wrong place was caused by me unintentionally ).

You mentioned l'Hopital. It is so to say your attempt at solution. What is it and what does it yield here ?

I doubt that is the exercise though, since the limit does not depend on n, assuming that there is no correlation between n and x.

Most likely, it is: ##\lim_{x \downarrow 0} \frac {8-9cos x+cos 3x} {sin^4(2x)}##

 

[Ray Vickson]

ok find limit as x→0 for the function ##[ 8-9cos x + cos 3x/{sin^4}2x]##
note that 2x is on the denominator...wish i could post this using latex...

It would be nice to post in LaTeX, but you can post clear and unambiguous questions simply by using parentheses properly! For example, here is your question in plain text, but using parentheses: (8 - 9 cos(x) + cos(3x))/sin^4(2x). Here is the same thing using in-line LaTeX: ##(8 - 9 \cos x + \cos 3x)/\sin^4(2x)##, and again using "displayed" |aTeX:
[tex] \frac{8 - 9 \cos x + \cos 3x}{\sin^4 2x} [/tex]
The first, plain-text version, is every bit as readable as the LaTeX versions, but of course it looks better. One advantage of the LaTeX versions is the need for fewer parentheses: ##\cos 3x## comes out perfectly clearly without the need for extra parentheses like ##\cos (3x)##. (However, I felt is necessary to use parentheses for clarity in ##\sin^4(2x)## in the in-line version but not in the displayed version.) They are all necessary in the plain text version, but you can sometimes make that easier to read by using different parenthesis style, as in [8 - 9 cos(x) + cos(3x)]/sin^4(2x).

 

[chwala]

Thanks sorry folks i have been on holiday...i will endeavour to look at this when i am free...

 

[chwala]

now using LHospital rule, we need to get the first derivative of
##\lim_{x \downarrow 0} \frac {8-9cos x+cos 3x} {sin^4(2x)}## because substituting 0 will give us an indeterminate form...on applying the rule i am getting:,
##\lim_{x \downarrow 0} \frac {9sin x-sin 3x} {8cos 2x.sin^3(2x)}## ok is this step correct?

 

[SammyS]

now using LHospital rule, we need to get the first derivative of
## \displaystyle \lim_{x \to 0} \frac {8-9\cos x+\cos 3x} {\sin^4(2x)}## because substituting 0 will give us an indeterminate form...on applying the rule i am getting:,
## \displaystyle \lim_{x \to 0} \frac {9\sin x-\sin 3x} {8\cos 2x\cdot\sin^3(2x)}## ok is this step correct?

First of all, you are not taking the derivative of the overall expression. Rather you are taking the derivatives of the numerator and denominator respectively.

You also have a error in the derivative of the numerator.

 

[chwala]

First of all, you are not taking the derivative of the overall expression. Rather you are taking the derivatives of the numerator and denominator respectively.

You also have a error in the derivative of the numerator.

but hospital rule says we take the derivatives indepedently i.e of the numerator and the denominator indepedently.Why should i take the derivative of the whole expression?

 

[SammyS]

but hospital rule says we take the derivatives indepedently i.e of the numerator and the denominator indepedently.Why should i take the derivative of the whole expression?

Right ! You should not take the derivative of the whole expression. But that's actually what you said you did in post #15 .

 

[chwala]

I made a slight mistake...
If ## f(x)= 8-9cos 2x + cos 3x ⇒ df/dx = 9 sin x - 3 sin 3x ##
If ##g(x)= {(sin 2x)^4} ⇒ dg/dx = 8cos 2x{(sin 2x)^3} ##
and this is lhopital's rule what do you mean that i took derivative of whole expression?

 

[member 587159]

I made a slight mistake...
If ## f(x)= 8-9cos 2x + cos 3x ⇒ df/dx = 9 sin x - 3 sin 3x ##
If ##g(x)= {(sin 2x)^4} ⇒ dg/dx = 8cos 2x{(sin 2x)^3} ##
and this is lhopital's rule what do you mean that i took derivative of whole expression?

He meant that you said in post #15 that you wanted to take the derivative of whole expression.

 

[chwala]

He meant that you said in post #15 that you wanted to take the derivative of whole expression.

ok are my derivatives in post 19correct?, if so how do i move from there...

 

[BvU]

You take the limit for ##x\downarrow 0## and see what comes out.

 

[member 587159]

ok are my derivatives in post 19correct?, if so how do i move from there...

df/dx is wrong. Try to find the correct derivative of that function.

 

[chwala]

df/dx is wrong. Try to find the correct derivative of that function.

##df/dx##? i don't understand...

 

[BvU]

f is short for numerator, like in: f/g.

 

[chwala]

You take the limit for ##x\downarrow 0## and see what comes out.

as x→0, we get 0, indeterminate form , meaning we should differentiate again?

 

[chwala]

df/dx is wrong. Try to find the correct derivative of that function.

posts 22 and 23 are contradicting...

 

[BvU]

No. I overlooked your error. MQ did not.

 

[SammyS]

I made a slight mistake...
If ## f(x)= 8-9\cos 2x + \cos 3x ⇒ df/dx = 9 \sin x - 3 \sin 3x ##
If ##g(x)= {(\sin 2x)^4} ⇒ dg/dx = 8\cos 2x{(\sin 2x)^3} ##
and this is lhopital's rule what do you mean that i took derivative of whole expression?

As Math_QED state there is a (small) error in your expression for ##\ df/dx \ .\ ## Likely it's a typo.

 

[BvU]

It's complicated helping you. What you wrote was

If ##f(x)= 8-9\cos 2x + \cos 3x ⇒ df/dx = 9 \sin x - 3 sin 3x##

What you meant to write was $$f(x)= 8-9\cos x + \cos 3x ⇒ df/dx = 9 \sin x - 3 \sin 3x$$

Now you have the derivative for the numerator and the derivative for the denominator. Both go to 0 for ##\ x\downarrow 0\ ##, so you end up with '0/0' and you have to do it again - and again

 

[chwala]

It's complicated helping you. What you wrote was

If ##f(x)= 8-9\cos 2x + \cos 3x ⇒ df/dx = 9 \sin x - 3 sin 3x##

What you meant to write was $$f(x)= 8-9\cos x + \cos 3x ⇒ df/dx = 9 \sin x - 3 \sin 3x$$

Now you have the derivative for the numerator and the derivative for the denominator. Both go to 0 for ##\ x\downarrow 0\ ##, so you end up with '0/0' and you have to do it again - and again

hahahahahahahhahaha BVU sometimes the brain goes off yeah you get the indeterminate form 0/0, let's do it all over again...thanks

 

[chwala]

now if ## f'(x) = 9 sin x-3sin 3x## , → ## f''(x) = 9cos x -9 cos 3x##
and
## g'(x) = 8cos 2x(sin 2x)^3 ##, →##g''(x)= -16(sin 2x)^4 + 96 (sin 2x)^2 (cos 2x)^2##
is this correct? substituting limits will still give 0/0

 

[BvU]

That's why I added the ... and again ...
If you do it right, the sequence will come to a satisfying end, and then we will hand you a quasi-alternative approach that works a little faster (and marginally less error prone as well).

 

[chwala]

That's why I added the ... and again ...
If you do it right, the sequence will come to a satisfying end, and then we will hand you a quasi-alternative approach that works a little faster (and marginally less error prone as well).

That's why I added the ... and again ...
If you do it right, the sequence will come to a satisfying end, and then we will hand you a quasi-alternative approach that works a little faster (and marginally less error prone as well).

is step 32 correct?

 

[BvU]

Haha, just suppose I am too lazy to check it and simply wait for the final answer - which I calculated yesterday on a piece of paper I accidentally discarded , Carry on !