Answer to Question #270753 in Calculus for Kamo

Question

For questions 1 consider the Maclaurin Series expansion of

x = + + + +1 x x2 x3 .....

e

2! 3!

1. The coefficient of the term containing x3 in the Maclaurin series expansion of e3x

is 3

1−e3x

Answer 1

$$e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...$$

1. Then

$$e^{3x}=1+3x+\dfrac{(3x)^2}{2!}+\dfrac{(3x)^3}{3!}+...$$

$$\dfrac{(3x)^3}{3!}=\dfrac{9}{2}x^3$$

The coefficient of the term containing $x^3$ in the Maclaurin series expansion of $e^{3x}$ is $9/2.$

2.

$$\lim\limits_{x\to0}\dfrac{1-e^{3x}}{x}$$

$$=\lim\limits_{x\to0}\dfrac{1-(1+3x+\dfrac{(3x)^2}{2!}+\dfrac{(3x)^3}{3!}+..)}{x}$$

$$=-\lim\limits_{x\to0}\dfrac{3x+\dfrac{(3x)^2}{2!}+\dfrac{(3x)^3}{3!}+..}{x}$$

$$=-\lim\limits_{x\to0}(3+\dfrac{9x}{2!}+\dfrac{27x^2}{3!}+...)=-3$$