Question

Does anyone know how I would solve this limit?

Answer 1

\[\lim_{x \rightarrow 0} \frac{ \sin(3x) }{ 2x }\]

Answer 2

Let u = 3x

Answer 3

\[
\lim_{x \rightarrow 0} \frac{ \sin(3x) }{ 2x } \\
u = 3x; ~~~x = \frac u3 \\
x \rightarrow 0 \implies u \rightarrow 0 \\
\lim_{x \rightarrow 0} \frac{ \sin(3x) }{ 2x } =
\lim_{u \rightarrow 0} \frac{ \sin(u) }{ 2u/3 } =
\lim_{u \rightarrow 0} \frac 32\frac{ \sin(u) }{ u } =
\frac 32 * \lim_{u \rightarrow 0} \frac{ \sin(u) }{ u } = ~~?\\
\]

Answer 4

The last part is a well-known limit:\[
\lim_{x \rightarrow 0} \frac{ \sin(x) }{ x } = 1
\]

Answer 5

Or, if you have been taught L'Hospital's Rule, you can differentiate the top and bottom of the original limit and then take the limit.