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

Answer 1

#"The Limit ="0.#

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#
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Answer 2

The limit is 0.

This limit can be evaluated in (at least) two ways. The first is by using l'Hopital's Rule, which states

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 using the identities of Pythagore, 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]}
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Answer 3

To evaluate the limit ( \lim_{\theta \to 0} \frac{1 - \cos(\theta)}{\sin(2\theta)} ), apply L'Hôpital's Rule:

[ \lim_{\theta \to 0} \frac{1 - \cos(\theta)}{\sin(2\theta)} = \lim_{\theta \to 0} \frac{d/d\theta(1 - \cos(\theta))}{d/d\theta(\sin(2\theta))} ]

[ = \lim_{\theta \to 0} \frac{\sin(\theta)}{2\cos(2\theta)} ]

Now, plug in ( \theta = 0 ) to get:

[ \frac{\sin(0)}{2\cos(0)} = \frac{0}{2 \cdot 1} = 0 ]

So, ( \lim_{\theta \to 0} \frac{1 - \cos(\theta)}{\sin(2\theta)} = 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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