Limit of Trig Function as x Approaches 0: Is the Answer 0/0?

In summary: Here the right side limit is 1 and the left side limit is -1. Since they are not equal, the limit does not exist.In summary, the limit of -sin(x)^2 / ( (sin(x)^2+x^3)*(cos(x)+1) ) as x approaches 0 is -1/2, while the limit of x/x = 1 as x approaches 0. However, the limit of ((x^2)^1/2) / (x + 2x^2) as x approaches 0 does not exist due to the right and left side limits not being equal.
  • #1
mathProb
13
0

Homework Statement



lim (cos x - 1) / (sin^2 x + x^3) as x approaches 0.

Homework Equations



sinx/x = 1

The Attempt at a Solution


I get 0/0. Is that the answer?
 
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  • #2
Yes, it has 0/0 form. That doesn't mean the limit is 0. Do you know l'Hopital's rule?
 
  • #3
Just a note: most of the limits you do will be in the form 0/0. What the limit actually equals to can be anything from negative infinity to positive infinity.
 
  • #4
I know l hospitals rule u take the derivative and then find the limit

but is there another way u can so it by using sinx/ x = 1
 
  • #5
mathProb said:
I know l hospitals rule u take the derivative and then find the limit

but is there another way u can so it by using sinx/ x = 1

Yes, there's another way. Start by multiplying numerator and denominator by cos(x)+1 and expand the numerator. Try it and see how far you get.
 
  • #6
cos^2 - 1 / ( sin^2 + x^3)(cos x + 1)
= - sin^2/ (sin^2 + x^3 )(cos +1)

is this the right way?
 
  • #7
i still get 0/0
 
  • #8
mathProb said:
cos^2 - 1 / ( sin^2 + x^3)(cos x + 1)
= - sin^2/ (sin^2 + x^3 )(cos +1)

is this the right way?

You are doing fine. Now divide numerator and denominator by sin(x)^2.
 
  • #9
- 1/cos + 1 + sin^2/x^3 cos x + sin^2/X^3
limit = 1?
 
  • #10
mathProb said:
- 1/cos + 1 + sin^2/x^3 cos x + sin^2/X^3
limit = 1?

You already weren't using enough parentheses to make it clear what you mean. Now you've lost all of them. I have no idea what you are trying to write. Start over and show the algebra steps you are using.
 
  • #11
(1/cos) + 1 + ?(sin^2/x^3 cos) + (sin^2/x^3)

limit =1
 
  • #12
mathProb said:
(1/cos) + 1 + ?(sin^2/x^3 cos) + (sin^2/x^3)

limit =1

http://en.wikipedia.org/wiki/Sigmoid_function

1d0e8454eabfab4cf9a14bb331954d70.png


[tex]x/x=1_{(\lim\rightarrow\infty)}[/tex]
 
  • #13
mathProb said:
(1/cos) + 1 + ?(sin^2/x^3 cos) + (sin^2/x^3)

limit =1

Start again from -sin(x)^2/((sin(x)^2+x^3)(cos(x)+1)). And explain step by step you got there. Divide numerator and denominator by sin(x)^2. There's no need to multiply the denominator out. What's lim x->0 of cos(x)+1?
 
  • #14
Dick said:
Start again from -sin(x)^2/((sin(x)^2+x^3)(cos(x)+1)). And explain step by step you got there. Divide numerator and denominator by sin(x)^2. There's no need to multiply the denominator out. What's lim x->0 of cos(x)+1?

Whs^

BTW I cut and pasted a png into my blog and got warned for it as a Homework Q?

Is that really necessary?

https://www.physicsforums.com/blog.php?b=1522

ie and who should I ask about it?

No offence but I don't think trig identities warrant that much attention do they?
 
Last edited by a moderator:
  • #15
Schrodinger's Dog said:
Whs^

BTW I cut and pasted a png into my blog and got warned for it as a Homework Q?

Is that really necessary?

https://www.physicsforums.com/blog.php?b=1522

ie and who should I ask about it?

No offence but I don't think trig identities warrant that much attention do they?

No idea. I can't see the blog entry anyway. You should take it up with the mentor who issued the warning. BTW, I don't see what sigmoids have to do with this problem. It's just a simple limits exercise.
 
Last edited by a moderator:
  • #16
Dick said:
No idea. I can't see the blog entry anyway. You should take it up with the mentor who issued the warning. BTW, I don't see what sigmoids have to do with this problem. It's just a simple limits exercise.

Sure np thanks Dick, helpful as ever and quick.

And congrats on being made a mod/homeworkhelper/Mentor btw, well deserved.
 
  • #17
1+ (sin^2/x^3(cosx+1))

limit of cos + 1 = 2
 
  • #18
mathProb said:
1+ (sin^2/x^3(cosx+1))

limit of cos + 1 = 2

Yes, the limit of cos(x)+1 is 2. As for "1+ (sin^2/x^3(cosx+1))", if that's supposed to be the denominator, it's still messed up. If you aren't going to show your algebra steps, I really can't tell you how you are messing up.
 
  • #19
-sin^2 / (sin^2 + X^3(cos +1))
= 1/(1+ (x^3(cos +1))/sin^2)
lim= 1
 
  • #20
That's -sin(x)^2 / ( (sin(x)^2+x^3)*(cos(x)+1) ). (cos(x)+1) multiplies both sin(x)^2 and x^3, not just one of them. So you get -1/( (1+x^3/sin(x)^2)*(1+cos(x)) ). Now what's the limit of x^3/sin(x)^2?
 
  • #21
lim = 0
 
  • #22
You should really speak in full sentences. If that's the answer to lim sin(x)^2/x^3 and you know why, then ok, so what's the limit of the full expression?
 
  • #23
lim = -1/2 as x approaches 0
 
  • #24
mathProb said:
lim = -1/2 as x approaches 0

Yes.
 
  • #25
Thanks a lot.
 
  • #26
How can a limit not exist?

limit of ((x^2)^1/2) / (x + 2x^2) as x approaches 0
= x / ( x (1+2x) )
=0 or dne
 
  • #27
The squareroot square can be interpreted as |x|

You then have
|x|/(x(1+2x))
The limit from the right side (assuming x > 0) is
lim 1/(1+2x) = 1
The limit from the left side (assuming x < 0) is
-1/(1+2x) = -1
Since the left and right side limits do not match, the limit does not exist.
 
  • #28
as x approaches 0?
 
  • #29
Yes, sorry.
Example: The function x/x = 1 everywhere except x=0 where it has a removable discontinuity (undefined). It's limit however as x goes to zero is 1.
[tex]\frac{x}{x(1+2x)}[/tex] will therefore have the same limit as
[tex]\frac{1}{1+2x}[/tex].
Your problem arises from the fact that the numerator is effectively |x| rather than just x.
The limit is defined only if the right and left sided limits are equal.
 

FAQ: Limit of Trig Function as x Approaches 0: Is the Answer 0/0?

What are the limits of trigonometric functions?

Trigonometric functions have a limit of infinity as the input approaches infinity or negative infinity. They also have a limit of negative infinity as the input approaches negative infinity. However, they do not have a limit at any specific value or when the input approaches a specific value.

What is the difference between the limits of sine and cosine functions?

The limits of sine and cosine functions are the same. As the input approaches infinity or negative infinity, both functions have a limit of infinity or negative infinity. However, the values of the functions at specific inputs may differ due to their different shapes and patterns.

Can the limit of a trigonometric function be undefined?

Yes, the limit of a trigonometric function can be undefined. This happens when the function has a vertical asymptote or a jump discontinuity at a specific input value. In this case, the limit does not exist as the input approaches that value.

How do you find the limit of a trigonometric function algebraically?

To find the limit of a trigonometric function algebraically, you can use the properties of limits and basic trigonometric identities. You can also use L'Hopital's rule if the limit results in an indeterminate form, such as 0/0 or infinity/infinity.

Are there any special cases when finding the limit of a trigonometric function?

Yes, there are some special cases when finding the limit of a trigonometric function. For example, if the input approaches a multiple of pi for sine or cosine, the limit may not exist. Also, if the input approaches 0 for tangent or cotangent, the limit may not exist due to the vertical asymptote at that value.

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