Which of the following is equal to limit $\lim _ { x \rightarrow 0 ^ { - } } - \frac { 2 x } { \sin ( 3 x ) }$ ?

A) $- \frac { 3 } { 2 }$
B) $- \frac { 2 } { 3 }$
C) $\frac { 2 } { 3 }$
D) $\frac { 3 } { 2 }$
E)Does not exist

Question

Which of the following is equal to limit $\lim _ { x \rightarrow 0 ^ { - } } - \frac { 2 x } { \sin ( 3 x ) }$ ?

A) $- \frac { 3 } { 2 }$
B) $- \frac { 2 } { 3 }$
C) $\frac { 2 } { 3 }$
D) $\frac { 3 } { 2 }$
E)Does not exist

Quizplus · Accepted Answer

The limit can be evaluated using the standard limit $\lim _ { x \rightarrow 0 } \frac { \sin x } { x } = 1$. By substituting $3x$ for $x$, we get $\lim _ { x \rightarrow 0 } \frac { \sin (3x) } { 3x } = 1$, which implies $\lim _ { x \rightarrow 0 } \frac { 3x } { \sin (3x) } = 1$. Therefore, the original limit becomes $\lim _ { x \rightarrow 0 ^ { - } } - \frac { 2x } { \sin (3x) } = - \frac { 2 } { 3 }$, as we multiply both numerator and denominator by $3$ to match the form of the standard limit.

Which of the Following Is Equal to Limit lim⁡x→0−−2xsin⁡(3x)\lim _ { x \rightarrow 0 ^ { - } } - \frac { 2 x } { \sin ( 3 x ) }limx→0−​−sin(3x)2x​

Which of the Following Is Equal to Limit $\lim _ { x \rightarrow 0 ^ { - } } - \frac { 2 x } { \sin ( 3 x ) }$