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

mathProb
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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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Yes, it has 0/0 form. That doesn't mean the limit is 0. Do you know l'Hopital's rule?
 
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.
 
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
 
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.
 
cos^2 - 1 / ( sin^2 + x^3)(cos x + 1)
= - sin^2/ (sin^2 + x^3 )(cos +1)

is this the right way?
 
i still get 0/0
 
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.
 
- 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


x/x=1_{(\lim\rightarrow\infty)}
 
  • #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?
 
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  • #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.
 
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  • #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.
\frac{x}{x(1+2x)} will therefore have the same limit as
\frac{1}{1+2x}.
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.
 
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