Solving Simple Trig Limit Homework

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In summary, the homework statement is trying to find a new limit for x such that the denominator is not 0. However, there is no way to factor the denominator out, and L'Hoptial's rule might be able to help.
  • #1
wvcaudill2
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Homework Statement


lim x->0 [tex](1-cosx)/sinx[/tex]


Homework Equations


[tex](sinx)/(1+cosx)[/tex] = [tex](1-cosx)/(sinx)[/tex]

[tex]sinx/x[/tex] = 1


The Attempt at a Solution


Well, I know I need to change my original limit, as I cannot let the denominator be 0. I usually try to factor out things, but there is nothing to factor out here. I'm not really sure what to do.
 
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  • #2
L'Hopital's rule might work well here. Have a look at that.
 
  • #3
JeSuisConf said:
L'Hopital's rule might work well here. Have a look at that.

Im afraid I don't know how to use this. I am in Calculus AB, so it is not something we learn. Any other ideas?

We usually factor, rationalize, etc. to find limits.
 
  • #4
If

[itex] \frac{1 - cos x}{sin x} = \frac{sin x}{1 + cos x} [/itex]

then, won't the limits of the two equations be equal?
 
  • #5
wvcaudill2 said:
Im afraid I don't know how to use this. I am in Calculus AB, so it is not something we learn. Any other ideas?

We usually factor, rationalize, etc. to find limits.

L'Hoptial's rule is extremely simple, and it makes solving the problem very obvious in this case. Anyways, with the equation you gave, the two limits will be the same as Strants said. You don't need any fancy work to evaluate that limit if you use that equation, because you can evaluate it exactly at the point x=0.
 
  • #6
Strants said:
If

[itex] \frac{1 - cos x}{sin x} = \frac{sin x}{1 + cos x} [/itex]

then, won't the limits of the two equations be equal?

I would think so, but when you graph this, it looks like there could be many limits.
 
  • #7
JeSuisConf said:
L'Hoptial's rule is extremely simple, and it makes solving the problem very obvious in this case. Anyways, with the equation you gave, the two limits will be the same as Strants said. You don't need any fancy work to evaluate that limit if you use that equation, because you can evaluate it exactly at the point x=0.



ok, thanks!
 

FAQ: Solving Simple Trig Limit Homework

What is a trigonometric limit?

A trigonometric limit is a mathematical concept used to describe the behavior of a trigonometric function as the input value approaches a specific value, often infinity or zero. It is used to determine the value of a limit at a point where a function may be undefined or take on an indeterminate form.

How do I solve a simple trigonometric limit?

To solve a simple trigonometric limit, you can use algebraic manipulations, trigonometric identities, and other basic limit-solving techniques. If the limit is in an indeterminate form, you can use L'Hopital's rule to evaluate it. It is also important to remember to simplify the expression as much as possible before plugging in the limit value.

What are some common trigonometric identities used to solve limits?

Some common trigonometric identities used to solve limits include the Pythagorean identities (sin²θ + cos²θ = 1), the sum and difference identities (sin(α ± β) = sinαcosβ ± cosαsinβ), and the double angle identities (sin2θ = 2sinθcosθ).

Can I use a calculator to solve trigonometric limits?

While a calculator can be a helpful tool in solving limits, it is important to remember that calculators can only provide an approximation of the limit and may not always give the exact answer. It is best to use a combination of algebraic and trigonometric techniques to solve limits by hand.

What are some real-life applications of trigonometric limits?

Trigonometric limits are used in a variety of fields, including physics, engineering, and astronomy. For example, in physics, limits are used to describe the behavior of particles as they approach the speed of light. In engineering, limits are used to calculate the stability of structures under different conditions. In astronomy, limits are used to determine the trajectory of celestial bodies.

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