Calculate # lim_(x rarr 0)x/(sin3x) =1/3 #?

1 Answer
Feb 8, 2018

# lim_(x rarr 0)x/(sin3x) =1/3 #

Explanation:

We seek:

# L = lim_(x rarr 0)x/(sin3x) #

# \ \ = 1/3 \ lim_(x rarr 0)(3x)/(sin3x) #

If we put # theta=3x# then as #x rarr 0 => theta rarr 0# and so we can write:

# L = 1/3 \ lim_(theta rarr 0)(theta)/(sin theta) #

And we know that a standard calculus limit is:

# lim_(theta rarr 0)(sin theta)/(theta) = 1 #

And so

# L = 1/3 \ 1/ (lim_(theta rarr 0)(sin theta)/(theta)) #
# \ \ = 1/3 #