What is the limit as x approaches 0 of #(2x)/tan(3x)#?

Question

Emma Aldridge · Accepted Answer

# lim_(x rarr 0) (2x)/tan(3x) = 2/3#

The limit: # lim_(x rarr 0) (2x)/tan(3x) # is of an indeterminate form #0/0#, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then, providing the limits exits then: # lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x)) # And so applying L'Hôpital's rule we get: # lim_(x rarr 0) (2x)/tan(3x) = lim_(x rarr 0) (2)/(3sec^2(3x))# # " "= 2/3lim_(x rarr 0) 1/(sec^2(3x))# # " "= 2/3#

Charles Arocha · Answer

To determine the limit graphically, se the graph below.

graph{2x/tan(3x) [-6.17, 6.316, -2.69, 3.547]} Zoom in (use wheel) until you can guess that the limit is about #0.667#

Ezra Adams · Answer

#2/3#

The first thing you should always try when calculating limits, is just entering the #x# value into the function:

#lim_{x \to 0} (2x)/(tan(3x)) = (2*0)/(tan(3*0)) = 0/tan(0)=0/0#

Seeing that plugging in gives an indeterminate form, we need to use L'Hospital's Rule.

#2lim_{x \to 0} (1)/((3csc^2(3x))/cot^2(3x))#

Note that I moved the 2 on top of the limit outside of the limit.

We can move the 3 in the denominator of the 2. #1/csc^2(x)# is #sin^2(x)# and #(1/1)/cot^2(x)# is simply #cot^2(x)#. But #cot^2(x)# is #cos^2(x)/sin^2(x)# so the #sin^2(x)# gets cancelled leaving:

#(2/3)lim_{x \to 0} cos^2(3x)#.

Cos(0)=0 so we're left with #2/3# which is our answer.

Ezra Adams · Answer

# lim_(x rarr 0) (2x)/tan(3x) = 2/3#

Another approach that doesn't rely on using L'Hôpital's rule We can write the limit as: # lim_(x rarr 0) (2x)/tan(3x) = lim_(x rarr 0) (2x)cot(3x)# # " " = 2lim_(x rarr 0) xcot(3x)# Let us look at the Taylor Series for #cotx#, which is as follows: #cotx=1/x-x/3-x^3/45 - ... # And so we can write a series expansion for the limit: # lim_(x rarr 0) (2x)/tan(3x) = 2lim_(x rarr 0) (x){1/((3x))-((3x))/3-(3x)^3/45 - ... }# # " "= 2lim_(x rarr 0) {1/3-x^2-(27x^4)/45 - ... }# # " "= 2(1/3)# # " "= 2/3#

Emma Aldridge · Answer

#lim_(xrarr0)(2x)/tan(3x) = 2lim_(xrarr0)(x/sin(3x) * cos(3x))# # 2lim_(xrarr0)(3x)/sin(3x) * lim_(xrarr0)cos(3x)# # = 2/3lim_(xrarr0)(3x)/sin(3x) * lim_(xrarr0)cos(3x)# # = 2/3(1)(1) = 2/3#

Aurora Alvarado · Answer

undefined The limit as x approaches 0 of (2x)/tan(3x) is 2/3.