Quick question about Trig limits (Calculus)

[JohnCU]

The limit as x -> 0 of (cosx - 1) / sinx...

I figure you'd break up the thing so it's like (cosx - 1 ) / x * x/sinx so you could say that x / sinx = 1 but cosx - 1/x = division by zero...

 

[Heisenberg]

Use L'Hoptial's rule. Take the derivative and you get -sin(x)/cos(x), as x ->0 gives you 0/1 = 0. I think.

 

[HokieESM]

L'Hopital's rule, my friend, L'Hopital's rule. 🙂

 

[JohnCU]

Can't use that, gotta do it by limit definition. 🙁

 

[Lyfer]

Originally posted by: JohnCU
The limit as x -> 0 of (cosx - 1) / sinx...

I figure you'd break up the thing so it's like (cosx - 1 ) / x * x/sinx so you could say that x / sinx = 1 but cosx - 1/x = division by zero...

Enuf of that, hurts my brain just thinking about it🙂.

 

[Excelsior]

Originally posted by: Lyfer

Originally posted by: JohnCU
The limit as x -> 0 of (cosx - 1) / sinx...

I figure you'd break up the thing so it's like (cosx - 1 ) / x * x/sinx so you could say that x / sinx = 1 but cosx - 1/x = division by zero...

Enuf of that, hurts my brain just thinking about it🙂.

Same here, and I am in calculus. What was I thinking when I signed up for it...

 

[HokieESM]

If you can't use L'Hopital's rule, multiply by (cosx+1)/(cosx+1). This will give:

((cosx)^2-1)/(sinx(cosx+1)) = -(sinx)^2/(sinx(cosx+1)) = -sinx/(cosx+1) = 0/2 = 0.

There you go. Answer also makes sense if you plot the original function..... OR use L'Hopital's rule.

 

[Kyteland]

Definition of Limit: http://mathworld.wolfram.com/Limit.html

So make a table:
x | eq(x)
-------------------
1 = -.5463
.1 = -.05
.01 = -.005
.001 = -.0005
.0001 = -.00005
.00001 = -.000005

Then try from the other direction

x | eq(x)
-------------------
-1 = .5463
-.1 = .05
-.01 = .005
-.001 = .0005
-.0001 = .00005
-.00001 = .000005

Looks to me like the limit is 0. It fits every criteria of the definition of a limit.

 

[Legendary]

I think maybe he means lim h -> 0 f(x+h) - f(x) / h

 

[JohnCU]

I think I got it now, thanks.