How do you evaluate #(1-cos2x)/(x^2)# as x approaches 0?

If we try to substitute 0 into #lim_(x\rightarrow 0)frac(1-cos2x) x^2#, we end up with #\frac0 0#. This introduces what is called an indeterminate form (there is another, infinity over infinity, but we don't have to worry about that one for this problem.)

When direct substitution yields an indeterminate form, we can use L'Hôpital's rule: #lim_(x\rightarrow a)frac(f(x)) (g(x))=lim_(x\rightarrow a)frac(f'(x)) (g'(x))#

Note that this is not the same as the Product Rule; rather, it means that the limit of the derivative of #f(x)# over the derivative of #g(x)# will be the same as the limit of #f(x)# over #g(x)#.

Let's try this:
#lim_(x\rightarrow 0)frac(1-cos2x) x^2=lim_(x\rightarrow 0)frac(2sin2x) 2x#.

What happens when we substitute 0 back into this?
#frac(2sin2(0)) (2(0))=\frac(2(0)) (0)#

We end up with #\frac0 0# again, but L'Hôpital's rule says we can do this as many times as necessary, so if we apply it again:
#lim_(x\rightarrow 0)frac(2sin2x) 2x =lim_(x\rightarrow 0)frac(4cos2x) 2#.

When we substitute 0 into this, we get
#\frac(4cos2(0)) (2)=\frac(4(1)) (2)#

This gives if #\frac4 2#, which reduces to #2#.

Thus, we know that #lim_(x\rightarrow 0)frac(1-cos2x) x^2# is #2#.

For more on L’Hôpital's rule, I encourage you to check out this link .

#lim_(xrarr0)(1-cos2x)/x^2 = lim_(xrarr0)(1-(1-2sin^2x))/x^2#

# = lim_(xrarr0)(2sin^2x)/x^2#

# = 2 (lim_(xrarr0)(sinx)/x)^2#

# = 2(1)^2 = 2#