What is the limit as x approaches 0 of #(2x)/tan(3x)#?
5 Answers
Explanation:
The first thing you should always try when calculating limits, is just entering the
Seeing that plugging in gives an indeterminate form, we need to use L'Hospital's Rule.
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.
Cos(0)=0 so we're left with
# lim_(x rarr 0) (2x)/tan(3x) = 2/3#
Explanation:
The limit:
# lim_(x rarr 0) (2x)/tan(3x) #
is of an indeterminate form
# 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#
# lim_(x rarr 0) (2x)/tan(3x) = 2/3#
Explanation:
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=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#
# 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#
To determine the limit graphically, se the graph below.
Explanation:
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