The discussion revolves around finding the limit of the expression (h sin h) / (1 - cos h) as h approaches 0, which falls under the subject area of limits in calculus, specifically involving trigonometric functions.
Some participants express agreement on the limit's value being 2, while others clarify the understanding of the limits involved. There is an acknowledgment of the approach being somewhat indirect, and one participant indicates a lack of familiarity with L'Hopital's Rule, suggesting that the current method is the only one they can use.
One participant notes that they have not yet learned L'Hopital's Rule, which may influence the methods available for solving the problem.
You're almost there!panadaeyes said:Homework Statement
Find the Limit: lim h->0 of ( h sin h)/ ( 1-cos h)
Homework Equations
lim h->0 of (sin theta)/ theta = 1
lim h->0 of (cos(h)-1)/h=0
The Attempt at a Solution
lim h->0 (h/1-cos h)*(sin h) * (1+cos h/ 1 +cos h)
= lim h->0 (h(1+cos h) * sin h)/(1-cos^2 h)
= lim h->0 (h(1+cos h))/sin h
That's how far I got, I have no idea how to continue.
Please and thank you!
Mark44 said:You're almost there!
[tex]=\lim_{h \to 0} \frac{h}{sin(h)}\cdot (1 + cos(h))[/tex]
You can break up the limit of a product to the product of limits, provided that both limits exist. Hint: look at your relevant equations.
Agreed.JonF said:Yup, I get 2 also.
Not sure what the missing symbol is - implies?JonF said:If you know limh→0 h/sinh = 1 then you know that
limh→0 (sinh/h)-1 = 1.
But (sinh/h)-1 = 1 ⇒ (sinh/h)= 1
But if the OP doesn't know L'Hopital's Rule, this is the way to go.JonF said:Edit: You don’t know l’hopital’s rule right? If you do this is a rather round-about way to do this problem.