How do you find the limit of #sin (x^2)/sin^2(2x)# as x approaches 0?

Question

Elijah Abram · Accepted Answer

#1/4#

We will be using

#lim_(x->0)sinx/x=1# so

#sin (x^2)/sin^2(2x)=((4x^2)/(4x^2))sin (x^2)/sin^2(2x)=1/4((sin(x^2)/x^2))/((sin(2x)/(2x))^2)# then

#lim_(x->0)sin (x^2)/sin^2(2x)=lim_(x->0)1/4((sin(x^2)/x^2))/((sin(2x)/(2x))^2)=1/4 1/1 = 1/4#

Paisley Johnson · Answer

undefined To find the limit of sin(x^2)/sin^2(2x) as x approaches 0, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately with respect to x.

Differentiating sin(x^2) with respect to x gives 2x*cos(x^2), and differentiating sin^2(2x) with respect to x gives 2*sin(2x)*cos(2x).

Now, we substitute x = 0 into the derivatives obtained above. We get 0 for both the numerator and denominator.

Next, we differentiate the derivatives obtained in the previous step.

Differentiating 2x*cos(x^2) gives 2*cos(x^2) - 4x^2*sin(x^2), and differentiating 2*sin(2x)*cos(2x) gives 4*cos^2(2x) - 4*sin^2(2x).

Again, we substitute x = 0 into these derivatives. We get 2 for the numerator and 4 for the denominator.

Finally, we divide the limit of the derivatives of the numerator and denominator, which is 2/4 = 1/2.

Therefore, the limit of sin(x^2)/sin^2(2x) as x approaches 0 is 1/2.