An interesting limit

D

Darya

Junior Member
Joined
Jan 17, 2020
Messages
154
Find a limit when x approaches 0 of :[MATH](1+sinx-cosx)/(1+sin6x-cos6x)[/MATH]The type of this limit is 0/0 so it'd be great to have product in both numerator and denominator but I'm quite lost here. I didnt find any identites I could use neither did writing 1 as sin^2x+cos^2x and doing manipulations with that help much. Any tips to tackle this?
Thanks for help. A lot.

Edit: It's not allowed to use L'Hopitals rule.
 
Hint: [math]\lim_{x->0}\dfrac{sin}{x}=1\ and \ \lim_{x->0}\dfrac{sin6x}{6x}=1[/math]
 
One thing you could try is to use the well known approximations

[MATH]x \to 0 \Rightarrow \sin(x) \approx x\\ x \to 0 \Rightarrow \cos(x) \approx 1[/MATH]
which is I guess what Jomo hinted at but in a less cryptic form.
 
Romsek said:
One thing you could try is to use the well known approximations

[MATH]x \to 0 \Rightarrow \sin(x) \approx x\\ x \to 0 \Rightarrow \cos(x) \approx 1[/MATH]
which is I guess what Jomo hinted at but in a less cryptic form.
Click to expand...
Romsek, exactly how will that help? If one replaces sin with x and cosx with 1 the limit is still 0/0. If your method works can you please show me how after the OP figures this out? Thanks!
 
Jomo said:
Romsek, exactly how will that help? If one replaces sin with x and cosx with 1 the limit is still 0/0. If your method works can you please show me how after the OP figures this out? Thanks!
Click to expand...

[MATH]\dfrac{1+\sin(x) -\cos(x)}{1 + \sin(6x)-\cos(6x)} \approx \dfrac{1+x - 1}{1 + 6x -1} = \dfrac{x}{6x} = \dfrac 1 6[/MATH]
Pretty simple eh?
 
Last edited:
Romsek said:
[MATH]\dfrac{1+\sin(x) -\cos(x)}{1 + \sin(6x)-\cos(6x)} \approx \dfrac{1+x - 1}{1 + 6x -1} = \dfrac{x}{6x} = \dfrac 1 6[/MATH]
Pretty simple eh?
Click to expand...
I've been away from teaching for way too long. I computed the [math]\lim{\dfrac{1+\sin(x) -\cos(x)}{1 + \sin(6x)-\cos(6x)}} \approx \lim\dfrac{1+x - 1}{1 + 6x -1}[/math] and got 0/0. After simplifying 1st the limit is obvious.
Any job offers out there?
 
I went with a transformation, just because this sort of transformation has always fascinated me.

[math]\dfrac{1-\sin(x)-\cos(x)}{1-\sin(6x)-\cos(6x)} = \dfrac{1-\sqrt{2}\cos\left(x-\dfrac{\pi}{4}\right)}{1-\sqrt{2}\cos\left(6x-\dfrac{\pi}{4}\right)}[/math]
This, of course, gets us to the same place.
 
To the OP: Whenever I see a limit as x->0 involving trig functions and direct substitution does not work then I always go for sinx/x or x/sinx (or some variation of that).
 
Jomo said:
To the OP: Whenever I see a limit as x->0 involving trig functions and direct substitution does not work then I always go for sinx/x or x/sinx (or some variation of that).
Click to expand...
Best advice, really. It's gonna help me enormously during the exam. I solved it this way:
Thank you so much.
 

Attachments

  • photo_2021-01-03_10-00-20.jpg
    photo_2021-01-03_10-00-20.jpg
    41.3 KB · Views: 11
Ouch, where did you learn you arithmetic from?

how can you pull the 1/x 's if front of the fraction?? Does [math]\dfrac{1+b -c}{1 + d - e}[/math] really equal [math] \dfrac{1}{1} + \dfrac{b - c}{d - e}[/math]
 
I guess that you did not pull out the 1's in front but it sure looked that way initially in your writing. My concern is why are you saying that (1-cosx)/x approaches 1/2 as x goes to 0. Can you please show me this proof? Remember do whatever you can to get a sinx over that x in the denominator
 
Jomo said:
I guess that you did not pull out the 1's in front but it sure looked that way initially in your writing. My concern is why are you saying that (1-cosx)/x approaches 1/2 as x goes to 0. Can you please show me this proof? Remember do whatever you can to get a sinx over that x in the denominator
Click to expand...
No, I didnt do that in my writing. I multiplied by (1/x)/(1/x), then also by (1/6)/(1/6), which is 1. I hope all other steps are clear, sorry for miscommunication. Proof for (1-cosx)/x is done by multiplying numerator and denominator by 1+cosx. We get sinx/x times 1/(cosx+x). The first limit is 1 and the second is 1/2. Is it okay?
 
Sorry but it is not OK. It seems that you are saying that (1-cosx)(1+cosx) = sinx while it is sin^2(x).

So you get (sin^2x)/(x*(1+cosx)) but the lim as x approaches 0 is NOT 1/2.

(sin^2x)/(x*(1+cosx)) = (sinx)(sinx/x)(1/(1+cosx)).
 
Jomo said:
Sorry but it is not OK. It seems that you are saying that (1-cosx)(1+cosx) = sinx while it is sin^2(x).

So you get (sin^2x)/(x*(1+cosx)) but the lim as x approaches 0 is NOT 1/2.

(sin^2x)/(x*(1+cosx)) = (sinx)(sinx/x)(1/(1+cosx)).
Click to expand...
*facepalm*
You are right. Yes, the limit of (1-cosx)/x^2 is 1/2. The limit "I've tried to prove" is zero. Thanks a lot.
 
You must log in or register to reply here.
Top