Master the Art of Finding Limits Involving Trig with Expert Homework Help

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Homework Help Overview

The discussion revolves around finding the limit of the expression involving trigonometric functions, specifically the limit as x approaches 0 of (tan(x))^2 / x. Participants express challenges with trigonometric concepts and limits in calculus.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss various approaches, including the potential use of L'Hôpital's rule and known limits involving sine and cosine. There are questions about the necessity of understanding trigonometric identities and how they relate to the problem at hand.

Discussion Status

There is an ongoing exploration of different methods to approach the limit problem. Some participants have begun to connect trigonometric identities to the limit, while others express confusion about the relationships between the functions involved. Guidance has been offered regarding the use of identities and known limits, but no consensus has been reached on a solution.

Contextual Notes

Participants note a lack of prior knowledge in trigonometry, which is impacting their ability to tackle the problem. There is mention of homework deadlines and the pressure to understand the material quickly, as well as references to the professor's teaching style and the pace of the course.

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Homework Statement


Finding limit: involves trig


Homework Equations



Lim X-> 0 ((tan(x))^2 / X)


The Attempt at a Solution



I've done multiple other problems, but can't do this one, my trig is weak.
 
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Torshi said:

Homework Statement


Finding limit: involves trig


Homework Equations



Lim X-> 0 ((tan(x))^2 / X)


The Attempt at a Solution



I've done multiple other problems, but can't do this one, my trig is weak.
Did you try L'Hôpital's rule ?
 
A well-known limit that can be used in this problem is this one:
$$ \lim_{x \to 0} \frac{sin(x)}{x} = 1$$
 
SammyS said:
Did you try L'Hôpital's rule ?

We're not there yet in lecture. I just started College level calculus. We have so far just defined what limits are and using εδ def. of limits and I'm having a hard time understanding the rules for using epsilon and delta to prove an argument and some limit problem solving involving trig etc.
 
Mark44 said:
A well-known limit that can be used in this problem is this one:
$$ \lim_{x \to 0} \frac{sin(x)}{x} = 1$$

I know this and the other one is = 0. My prof said memorize it, but I don't know how to incorporate it. I feel like I'm in a rough start for calc...I'm not bad in math per se, I've taken college alg, statistics, physics 1/2, gen chem 1/2 etc all a year ago or more, haven't taken trig or precal. I took the math placement and scored enough somehow to end up in this hard class that i need.
 
Torshi said:
I know this and the other one is = 0. My prof said memorize it, but I don't know how to incorporate it. I feel like I'm in a rough start for calc...I'm not bad in math per se, I've taken college alg, statistics, physics 1/2, gen chem 1/2 etc all a year ago or more, haven't taken trig or precal. I took the math placement and scored enough somehow to end up in this hard class that i need.
Write tan(x) in terms of sin(x) & cos(x) .
 
SammyS said:
Write tan(x) in terms of sin(x) & cos(x) .

are you implying i should need to know the identities such as 1-cosx = sinx^2 for example - not that this has to do anything with the prob? Sorry if I'm off, but I never took trig
 
Torshi said:
are you implying i should need to know the identities such as 1-cosx = sinx^2 for example - not that this has to do anything with the prob? Sorry if I'm off, but I never took trig

You need to know tan(x)=sin(x)/cos(x). You do need to know a LITTLE trig.
 
Torshi said:
are you implying i should need to know the identities such as 1-cosx = sinx^2 for example
It's actually cos2(x) = 1 - sin2(x).
Torshi said:
- not that this has to do anything with the prob? Sorry if I'm off, but I never took trig
That's something you'll need to rectify, since many of the problems you'll see will require some knowledge of basic trig identities. For starters, you need to be able to write tan(x), cot(x), sec(x), and csc(x) in terms of sin(x) and/or cos(x).
 
  • #10
Dick said:
You need to know tan(x)=sin(x)/cos(x). You do need to know a LITTLE trig.

I still can't solve it. I know sin(x)/x = 1

edit: Alright, I'll brush up on my trig identities etc. But, this HW is due on Tuesday. I'll youtube how to do that
 
  • #11
Torshi said:
I still can't solve it. I know sin(x)/x = 1
No, ##lim_{x \to 0} \frac{sin(x)}{x} = 1##, but that's different from what you wrote.
Torshi said:
edit: Alright, I'll brush up on my trig identities etc. But, this HW is due on Tuesday.
 
  • #12
So I have written down all the identities. Would (Tan(x)^2) which is the numerator first turn into Tan^2(x) which i believe is another way of writing it.

I also have down here that Tanθ = Sinθ / Cosθ.....Also tan^2θ = Sec^2θ-1 ?

Edit: I did a similar problem and got it right. lim X--> 0 F(x) = 1-cosx / x

1.) 1-cosx / x multiplied by 1+cosx / 1+cosx
2.) 1^2 - cosx^2 / x (1+cosx) then becomes sin^2x / x(1+cosx)
3.) 1 x 0 = 0
 
Last edited:
  • #13
So far I'm assuming I have:

1.) tan^2x = Sec^2x - 1

2.) Sec^2x -1 / x X Sec^2x+1 / Sec^2+1

3.) Sec^4 -1 / x (sec^2x+1)

?? good so far?

edit: I think I'm doing it wrong
 
  • #14
Go back to tan = sin/cos, sin(x)/x tends to 1 as x tends to zero. What do you think cos(x) tends to as x tends to 0?
 
  • #15
haruspex said:
Go back to tan = sin/cos, sin(x)/x tends to 1 as x tends to zero. What do you think cos(x) tends to as x tends to 0?

I think the answer is zero I believe for my overall problem
 
  • #16
Yes, but why?
 
  • #17
haruspex said:
Yes, but why?

I don't know I'm so confused, this crap has been bugging me all day. It's been a while since I've done all this.

All my prof said was that lim x-->0 sinx/x = 1 and lim x-->0 1-cosx/x = 0 . This is calc so he just assumes we know it, he didn't review any precalc or trig obv. He just speeds right through.

I wrote down all the trig identities such as tanx=sinx/cosx etc etc and cos^2x + sin^2x =1

I know these things are all related. In one of my posts I figured out an answer due to trig identities. I can't for this one... I'm sure it's very simple, I just can't see it.

I don't see how I can relate (tan(x)^2)/x to any of them...
 
  • #18
Torshi said:
I think the answer is zero I believe for my overall problem

Yes, it is. Can you explain why?
 
  • #19
Torshi said:
I don't know I'm so confused, this crap has been bugging me all day. It's been a while since I've done all this.

All my prof said was that lim x-->0 sinx/x = 1 and lim x-->0 1-cosx/x = 0 . This is calc so he just assumes we know it, he didn't review any precalc or trig obv. He just speeds right through.

I wrote down all the trig identities such as tanx=sinx/cosx etc etc and cos^2x + sin^2x =1

I know these things are all related. In one of my posts I figured out an answer due to trig identities. I can't for this one... I'm sure it's very simple, I just can't see it.

I don't see how I can relate (tan(x)^2)/x to any of them...

tan(x)^2/x=(sin(x)/x)*(sin(x)/cos(x)^2). What's the limit of each of those two factors?
 
  • #20
Torshi said:
I don't know I'm so confused, this crap has been bugging me all day. It's been a while since I've done all this.

All my prof said was that lim x-->0 sinx/x = 1 and lim x-->0 1-cosx/x = 0 .
That had better be lim x-->0 (1-cosx)/x = 0 .
I wrote down all the trig identities such as tanx=sinx/cosx etc etc and cos^2x + sin^2x =1
Do you know what cos(0) is? Given that, and knowing that sin(x)/x -> 0, and sin(x) = cos(x)tan(x), what can you say about lim x->0 tan(x)/x?
 
  • #21
Dick said:
tan(x)^2/x=(sin(x)/x)*(sin(x)/cos(x)^2). What's the limit of each of those two factors?

That would be 1 X 0 = 0 ?but I don't understand how tan(x)^2/x = what you posted above. .
 
Last edited:
  • #22
haruspex said:
That had better be lim x-->0 (1-cosx)/x = 0 .

Do you know what cos(0) is? Given that, and knowing that sin(x)/x -> 0, and sin(x) = cos(x)tan(x), what can you say about lim x->0 tan(x)/x?

cos(0) = 1?
 
  • #23
Torshi said:
That would be 1 X 0 = 0


but I don't understand how tan(x)^2/x = what you posted above. .

tan(x)=sin(x)/cos(x). So tan(x)^2/x=sin(x)^2/(cos(x)^2*x)=(sin(x)/x)*(sin(x)/cos(x)^2). Just regroup the factors.
 
  • #24
Torshi said:
... I know tanx = sinx/cosx.
Right.

So if \displaystyle \ \ \tan(x)=\frac{\sin(x)}{\cos(x)}\,, \ then \displaystyle \ \ \frac{\tan^2(x)}{x}=\frac{\sin(x)\tan(x)}{x\cos(x)}\ . \ Right?

Of course that can be written \displaystyle \ \ \frac{\sin(x)}{x}\frac{\tan(x)}{\cos(x)}\ . \

Does that help ?
 
  • #25
Dick said:
tan(x)=sin(x)/cos(x). So tan(x)^2/x=sin(x)^2/(cos(x)^2*x)=(sin(x)/x)*(sin(x)/cos(x)^2). Just regroup the factors.

Thank god I get it now! Jesus lol
 
  • #26
SammyS said:
Right.

So if \displaystyle \ \ \tan(x)=\frac{\sin(x)}{\cos(x)}\,, \ then \displaystyle \ \ \frac{\tan^2(x)}{x}=\frac{\sin(x)\tan(x)}{x\cos(x)}\ . \ Right?

Of course that can be written \displaystyle \ \ \frac{\sin(x)}{x}\frac{\tan(x)}{\cos(x)}\ . \

Does that help ?
Yes! OMg
 
  • #27
Thank you!
 
  • #28
Actually, just before SammyS's post Dick had told you that
\frac{tan^2 x}{x}= \frac{sin(x)}{x}\frac{1}{cos(x)}\frac{sin(x)}{cos(x)}
which, I think, is a simpler approach.
 

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