calc problem?

Question

lim ((cos (2x) - 1)/(x^2))
x--> 0

Answer 1

Ok, let's go on :

lim ((cos (2x) - 1)/(x^2))
x--> 0

Let's apply the rule of L'Hospitall :

We need to derivative the numerator and denominator, until we can replace x = 0

Lim (( -2*sin(2x) / 2x))
x-->0

Lim(( -4*cos(2x) / 2))
x-->0

Now, we can replace x = 0

Lim ((-4 / 2))

So the result will be = -2

The limit = -2

That's it

Answer 2

Use L'Hopital's rule twice. If th elimit takes the form 0/0 (which this one does), take the derivative of the top, the derivative of the bottom, and try the limit again. Again, you get 0/0, so take the derivatives AGAIN. This time, you should be able to evaulate the limit by simply plugging in x = 0.


EDIT: Niftier way. Since cos(2x) = 1-2sin^2 x, you can rewrite the limit as

lim (-2sin^2 x)/(x^2),
x-->0

or -2 [ lim (sin x / x) ]^2


That limit on the inside should look familiar.

Answer 3

Answer : -2


0/0 indefinite
take derivative numerator and denumenator

lim x-->0 -2sin2x/2x
=-2*1=-2

Answer 4

Graphically, the limit is -2 as you approach from x>0

Answer 5

lim ((cos (2x) - 1)/(x^2))
x--> 0
=lim (-2sin(2x))/2x
x --> 0
= lim -4cos(2x)/2
x --> 0
= -4/2 = -2