evaluate limx->0(1-cosx)/2x^2?

Answer 1

the answer is zero

zero multiplied or divided by anythin is always zero !!


therefore the above relation will work out to be zero.

Answer 2

limx->0 (1 - cosx) / 2x^2

Use L'hopital's rule and take derivatives of both numerator and denominator, until the limit evaluates to a number:

1) limx->0 (1 - cosx) / 2x^2 can't evaluate
2) limx->0 (sinx) / 4x can't evaluate
3) limx->0 (cosx) / 4 CAN evaluate

Plug in 0 for x:

(cos(0)) / 4
=(1) / 4
=1/4

limx->0 (1 - cosx) / 2x^2 = 1/4

Answer 3

The answer is 1/4. However you don’t “have” to use l’Hopital’s rule (although it is quickest).

The series for cos(x) is:

Cos(x) = ∑ (-1)^n x^(2n)/(2n!) where the sum is from n = 0 to n = infinity

= 1 – x^2/2 + x^4/4! – x^6/6! + …

So if f(x) = (1 – cos(x))/(2x^2)

Then 1 – cos(x) = x^2/2! - x^4/4! + x^6/6! - …

Divide by 2x^2 term by term and now take the limit.

You should see that only the first term (x^2/2)/(2x^2) = 1/4 survives the limit, the rest all equal zero.

Again, l’Hopital’s rule is quickest, but you don’t "have" to use it.

Answer 4

It's 1/4:

=limx->0 (sinx/4x)=limx->0(cosx/4)=1/4.

L'Hopitalle (or something like that) rule.

Answer 5

limx->0(1-cosx)/2x^2?
=limx->0{sin^2(x/2)}/2x^2 Because cos x =1 -2sin^2(x/2)
=limx->0{sin^2(x/2)}/x^2
=limx->0{sin^2(x/2)}/4(x/2}^2
Taking the limits we get
=1/4 Because limx->0 (sinx?x)=1

Answer 6

The answer is 1/4. You need to use L'Hopitals rule, twice. First time you will get sin(x)/4x. Second time around it will be cos(x)/4. Then plug in 0 to get cos(0)/4 = 1/4. FYI, the reason you have to use L'Hopitals rule is because you can't divide by zero. Happy days.

Answer 7

0 everybody knows that 0 times something is still 0

Answer 8

X is a varible and this is not an equation! You have to solve for X