Which of the following is true for the limit $\lim _ { x \rightarrow 0 } \frac { e ^ { 2 x } - 1 } { 1 - e ^ { 3 x } }$ ?

A) $- \frac { 2 } { 3 }$
B) $- \frac { 3 } { 2 }$
C)0
D) $\frac { 2 } { 3 }$
E) $\frac { 3 } { 2 }$

Question

Which of the following is true for the limit $\lim _ { x \rightarrow 0 } \frac { e ^ { 2 x } - 1 } { 1 - e ^ { 3 x } }$ ?

A) $- \frac { 2 } { 3 }$
B) $- \frac { 3 } { 2 }$
C)0
D) $\frac { 2 } { 3 }$
E) $\frac { 3 } { 2 }$

Quizplus · Accepted Answer

The limit can be solved by applying L'Hôpital's Rule, which is used when a limit is in an indeterminate form like 0/0. Differentiating the numerator and the denominator separately with respect to $x$ gives: $\lim _ { x \rightarrow 0 } \frac { 2e ^ { 2 x } } { -3e ^ { 3 x } }$. Evaluating this at $x = 0$ gives $\frac { 2 } { -3 }$, which simplifies to $- \frac { 2 } { 3 }$.

Which of the Following Is True for the Limit lim⁡x→0e2x−11−e3x\lim _ { x \rightarrow 0 } \frac { e ^ { 2 x } - 1 } { 1 - e ^ { 3 x } }limx→0​1−e3xe2x−1​

Which of the Following Is True for the Limit $\lim _ { x \rightarrow 0 } \frac { e ^ { 2 x } - 1 } { 1 - e ^ { 3 x } }$