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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?
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
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?
mathProb said:- 1/cos + 1 + sin^2/x^3 cos x + sin^2/X^3
limit = 1?
mathProb said:(1/cos) + 1 + ?(sin^2/x^3 cos) + (sin^2/x^3)
limit =1
mathProb said:(1/cos) + 1 + ?(sin^2/x^3 cos) + (sin^2/x^3)
limit =1
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?
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?
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.
mathProb said:1+ (sin^2/x^3(cosx+1))
limit of cos + 1 = 2
mathProb said:lim = -1/2 as x approaches 0
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.
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.
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.
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.
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.