Determine the value of the limit $\lim _ { x \rightarrow 0 } \frac { x ^ { 3 } } { \sin x - x }$ ?
A) $- \infty$
B) $- 6$
C)6
D)0
E) $\infty$

Question

Quizplus · Accepted Answer

The limit can be evaluated using L'Hôpital's Rule, which states that if the limit of a function is in an indeterminate form (like 0/0 or ∞/∞), then the limit of the function can be found by taking the derivative of the numerator and the derivative of the denominator separately and then taking the limit of the new function. Applying L'Hôpital's Rule three times to the given function, we get:1st application: $\lim _ { x \rightarrow 0 } \frac { 3x^2 } { \cos x - 1 }$2nd application: $\lim _ { x \rightarrow 0 } \frac { 6x } { -\sin x }$3rd application: $\lim _ { x \rightarrow 0 } \frac { 6 } { -\cos x }$Evaluating the final expression at $x = 0$, we get $\frac { 6 } { -1 } = -6$, hence the correct answer is B.

Determine the Value of the Limit lim⁡x→0x3sin⁡x−x\lim _ { x \rightarrow 0 } \frac { x ^ { 3 } } { \sin x - x }limx→0​sinx−xx3​

Determine the Value of the Limit $\lim _ { x \rightarrow 0 } \frac { x ^ { 3 } } { \sin x - x }$