Question

Evaluate the limit: `lim_(theta-> 0)(1-cos theta)/(sin 2theta)`?

Answer 1

`"The Limit ="0.`

Explanation:

We will use the Identity `: sin2theta=2sinthetacostheta`

Note that, `(1-costheta)/(sin2theta)`

`={(1-costheta)/(sin2theta)}{(1+costheta)/(1+costheta)}`

`=(1-cos^2theta)/{(sin2theta)(1+costheta)}`

`=sin^2theta/{(2sinthetacostheta)(1+costheta)}`

`=(sintheta)/(2costheta(1+costheta))`

`=(sin0)/{(2)(cos0)(1+cos0)}=0/4`

`:." the Reqd. Lim.="0`

Answer 2

The limit is 0.

Explanation:

There are (at least) two ways to evaluate this limit. The first way is to apply l'Hopital's Rule, which says

If both `f(x) -> 0 " and " g(x) -> 0 " as " x -> 0`

then `lim_(x->0)(f(x))/(g(x))" "=" "lim_(x->0) (f'(x))/(g'(x))`

Since both `1-cos theta` and `sin 2theta` approach 0 as `theta -> 0`, we can apply l'Hopital's rule to get:

`color(white)= lim_(theta->0) (1-cos theta)/(sin 2theta)`

`=lim_(theta->0)sintheta/(2cos(2theta))`

After this step, we no longer have a "division by zero" limit, so we can just plug in 0 for `theta` to get our answer:

`= sin(0)/(2cos[2(0)])`

`= 0/(2(1))`

`= 0`

Another way possible is to use `sin 2theta = 2sintheta costheta`, and also multiply both numerator and denominator by `1+costheta` to achieve

`color(white)= lim_(theta->0) (1-cos theta)/(sin 2theta)`

`=lim_(theta->0) (1-cos theta)/(2sin theta cos theta) * (1+ cos theta)/(1 + costheta)`

`=lim_(theta->0) (1-cos^2 theta)/(2 sin theta cos theta (1 + costheta))`

By the Pythagorean identities, we have

`=lim_(theta->0) (sin^2 theta)/(2 sin theta cos theta (1 + costheta))`

`=lim_(theta->0) (sin theta)/(2 cos theta (1 + costheta))`

Once again, we no longer have a "division by zero", so we can directly substitute 0 for `theta` to get

`=(sin (0))/(2 cos (0) [1 + cos(0)])`

`=(0)/(2(1)[1 + 1])`

`=0`.

Here is a graph of the function `f(theta)=(1-cos theta)/(sin 2theta)`:
graph{(1-cos x)/(sin(2x)) [-10, 10, -5, 5]}