How do you find the limit of #((sin^2)3x) / (x^2)# as x approaches 0?

Question

Emily Ammerman · Accepted Answer

#=9#

we will use well-known limit #lim_(z to 0) (sin z)/z = 1#. We need to do some manipulation to use it here

#lim_(x to 0) (sin^2 3x) / (x^2)#

#=lim_(x to 0) (sin 3x) / (x) (sin 3x) / (x)#

#=lim_(x to 0)3 (sin 3x) / (3x) 3 (sin 3x) / (3x)#

and we factor out the constants #=9 lim_(x to 0) (sin 3x) / (3x) (sin 3x) / (3x)#

with sub #z = 3x#, we have

#=9 lim_(z to 0) (sin z) / (z) (sin z) / (z)#

and the limit of the product is the product of the limits

#=9 lim_(z to 0) (sin z) / (z) lim_(z to 0) (sin z) / (z)#

#=9* 1* 1 = 9#

Ezekiel Alexander · Answer

undefined To find the limit of ((sin^2)3x) / (x^2) as x approaches 0, we can use the limit properties and trigonometric identities. By applying the limit properties, we can rewrite the expression as (sin^2(3x)) / (x^2) = (sin(3x))^2 / (x^2).

Next, we can use the trigonometric identity lim(x→0) sin(x) / x = 1. By substituting 3x for x, we have lim(x→0) sin(3x) / (3x) = 1.

Using this identity, we can rewrite the expression as lim(x→0) (sin(3x))^2 / (x^2) = (lim(x→0) sin(3x) / (3x))^2 = 1^2 = 1.

Therefore, the limit of ((sin^2)3x) / (x^2) as x approaches 0 is equal to 1.