How do you solve lim x->0 (sin(1 - cos x)) / x?

Answer 1

lim(x->0) [sin(1-cos x)/x] = sin(1-cos 0)/0 = 0/0
Since we got the indeterminite form 0/0, use L'Hopital's rule and take the derivatives of the top and bottom:
lim(x->0) [sin(1-cos x)/x] = lim(x->0) [(sin x)cos(1-cos x)/1]
(sin 0)cos(1 - cos 0) = 0*cos 0 = 0

The limit is 0.

Answer 2

Step 1. Plug in x=0 into sin(1-cos x)/x to get 0/0 which is an indeterminate. This makes it qualify to be solved by L'hopital's rule.

Step 2. apply L'hopital's, by taking the derivative of the numerator and denominato:
lim x->0 d/dx[sin(1-cos x)]/d/dx(x)
=lim x->0 [cos(1-cos x) sin x]/1. Note do not forget to use chain rule in the numerator.

Step 3. Plug in x=0 in the last expression to get:
cos (1-cos 0)sin 0=cos (1-1)sin 0= 0.
Thus the answer is 0.

Answer 3

It easy.... Just we use that lim x->0 (1 - cos x) / x = 0. Multiply and divide by 1-cos x because for a given small 'x' non zero, 1-cos x is non zero.

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

= lim x->0 ( (1 - cos x)/x ) * sin(1 - cos x) / (1 - cos x)

Then, using lim x->0 sen(x)/x = 1, we have

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

= lim x->0 ( (1 - cos x)/x ) * lim x->0 sin(1 - cos x) / (1 - cos x)
= 0 * 1
=0.

....

Answer 4

lim x->0 cos(x)/x DNE lim (x-)->0 cos(x)/x = -(cos(0)/0) = -a million/0 = -? lim (x+)->0 cos(x)/x = (cos(0)/0) = a million/0 = ? The reduce itself would not exist. also, in case you propose lim x->0 (cos(x/x)) then it may only be cos(a million). the graph of cos(x/x) is a without delay horizontal line that has a hollow at x=0.

Answer 5

The limit is 0/0 so you can use L'Hospital's Rule: lim f/g = lim f'/g'.

So this equals lim x->0 (cos(1 - cos x) * sinx) / 1
= cos(1-cos)) * sin0
= 0.

Answer 6

at limit plug in value

cos (0)= 1
so sin(1-0)= sin(1)

but numerator x=0

so lim x=0 = 1/0 or infinity

Answer 7

Do your own homework.