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

5 Answers

2/3

Explanation:

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.

Feb 18, 2017

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

Explanation:

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

Feb 18, 2017

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, 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

Feb 19, 2017

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

Feb 19, 2017

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 0.667