Use series to evaluate the limit.
-$$\lim _ { x \rightarrow 0 } \frac { e ^ { 4 x } - 1 - 4 x } { x ^ { 2 } }$$
A) 16
B) 4
C) 8
D) 0

Question

Quizplus · Accepted Answer

The given limit can be evaluated using the Maclaurin series expansion for $e^{4x}$, which is $1 + 4x + \frac{(4x)^2}{2!} + \frac{(4x)^3}{3!} + \cdots$. Substituting this into the expression and simplifying, the terms $1$ and $4x$ cancel out, leaving $\frac{16x^2}{2} + $ higher order terms, which simplifies to $8x^2$ plus higher order terms in the numerator. Dividing by $x^2$ and taking the limit as $x$ approaches 0 leaves us with $8$, as the higher order terms become negligible.

Use Series to Evaluate the Limit lim⁡x→0e4x−1−4xx2\lim _ { x \rightarrow 0 } \frac { e ^ { 4 x } - 1 - 4 x } { x ^ { 2 } }x→0lim​x2e4x−1−4x​

Use Series to Evaluate the Limit $\lim _ { x \rightarrow 0 } \frac { e ^ { 4 x } - 1 - 4 x } { x ^ { 2 } }$