How do I find the limit of x/(sin3x)?

1 Answer
Mar 7, 2018

"The answer is:" \qquad \qquad \qquad lim_{ x rarr 0 } x/sin(3x) \ = \ 1/3.

Explanation:

"I assume you mean limit, as" \ x \ "goes toward 0 (?). We can"
"proceed as follows. The idea is to try to take advantage of"
"the Fundamental Trig Limit:" \quad lim_{ A rarr 0 } sin(A)/A \ = \ 1."

"Proceeding:"

\qquad lim_{ x rarr 0 } x/sin(3x) \ = \ lim_{ x rarr 0 } 1/{ sin(3x)/x }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ lim_{ x rarr 0 } 1/{ 3/3 cdot sin(3x)/x }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ lim_{ x rarr 0 } 1/{ 3 cdot [ sin(3x)/{ 3 x } ] }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ lim_{ x rarr 0 } 1/3 cdot 1/{ [ sin(3x)/ { 3 x } ] }

color{blue}{ "now use the Fundamental Trig Limit:" \quad lim_{ A rarr 0 } sin(A)/A \ = \ 1; }
color{blue}{ \qquad "and use it with" \ A \ = \ 3 x. \ \ "Continuing:" }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ 1/3 cdot 1/{ [1] }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ 1/3.

"This is our answer:"

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad lim_{ x rarr 0 } x/sin(3x) \ = \ 1/3.