How do I solve #lim_(x to 0) (3x^4)/(sin^2(3x^2))#?

1 Answer
Steve M
Feb 22, 2018

# lim_(x to 0) (3x^4)/(sin^2(3x^2)) = 1/3#

Explanation:

We seek:

# L = lim_(x to 0) (3x^4)/(sin^2(3x^2))#

Which we can write as follows:

# L = 3 \ lim_(x to 0) (3x^2)/(sin(3x^2)) * (3x^2)/(sin(3x^2)) * 1/9#

# \ \ = 1/3 \ lim_(x to 0) (3x^2)/(sin(3x^2)) * \ lim_(x to 0) (3x^2)/(sin(3x^2)) #

The we use the standard calculus limit:

# lim_(theta to 0) sinx/x=1 #

to get:

# L = 1/3 * 1 * 1 #
# \ \ = 1/3 #

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