# 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