Calculus Help

[MrCodeDude]

First of all, thanks in advance.

Determine the limit of the trigonometric function (if it exists)

limit as x -> 0

(tan^2 x) / x

And, just for your information:
(sin x) / x = 1 and (1 - cos(x)) / x = 0

 

[Somecallmetim]

Im sorry if I'm wrong, but since when can Tangent be squared?
Do you mean Tan x^2?

 

[notfred]

Why the hall can't tangent be squared? tan^2 x = (tan x)^2

 

[Legendary]

Use L'Hospital's rule and take the derivative of the top and the bottom then reapply the limit

 

[phatj]

L'Hospital's rule....

lim as x approaches 0 of (2*tan(x)*sec^2(x))/1

 

[phatj]

or if you are not allowed to use L'Hospital's rule (or havent learned it yet)

do this:

lim x -> 0 of [tan^2(x)/x] = lim x -> 0 of [sin(x) * [sin(x)/x] * 1/(cos^2(x))] = sin(0) * 1 * 1/1^2 = 0

remember: tan(x) = sin(x)/cos(x), so tan^2(x) = sin^2(x)/cos^2(x)

 

[Syringer]

Yeah, was about to post that..convert tan to sin/cos, cos is left in the denominator, and the limit to 0 makes the denominator one..