Question

use series to evaluate the limit.

$$\\lim _ { x \ ightarrow 0 } \\frac { \\sin ( 4 x ) - 4 x + \\frac { 32 } { 3 } x ^ { 3 } } { x ^ { 5 } }$$

Explanation:

Step1: Recall the Maclaurin series for $\sin t$

The Maclaurin series for $\sin t=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{(2n+1)!}t^{2n + 1}=t-\frac{t^{3}}{3!}+\frac{t^{5}}{5!}-\frac{t^{7}}{7!}+\cdots$
Let $t = 4x$, then $\sin(4x)=(4x)-\frac{(4x)^{3}}{3!}+\frac{(4x)^{5}}{5!}-\frac{(4x)^{7}}{7!}+\cdots=4x-\frac{64x^{3}}{6}+\frac{1024x^{5}}{120}-\frac{16384x^{7}}{5040}+\cdots=4x-\frac{32x^{3}}{3}+\frac{128x^{5}}{15}-\frac{1024x^{7}}{315}+\cdots$

Step2: Substitute the series into the limit expression

Substitute $\sin(4x)=4x-\frac{32x^{3}}{3}+\frac{128x^{5}}{15}-\frac{1024x^{7}}{315}+\cdots$ into $\lim_{x
ightarrow0}\frac{\sin(4x)-4x+\frac{32x^{3}}{3}}{x^{5}}$

We get $\lim_{x
ightarrow0}\frac{(4x-\frac{32x^{3}}{3}+\frac{128x^{5}}{15}-\frac{1024x^{7}}{315}+\cdots)-4x+\frac{32x^{3}}{3}}{x^{5}}$

Simplify the numerator: $(4x - 4x)+(-\frac{32x^{3}}{3}+\frac{32x^{3}}{3})+\frac{128x^{5}}{15}-\frac{1024x^{7}}{315}+\cdots=\frac{128x^{5}}{15}-\frac{1024x^{7}}{315}+\cdots$

Step3: Evaluate the limit

Now, $\lim_{x
ightarrow0}\frac{\frac{128x^{5}}{15}-\frac{1024x^{7}}{315}+\cdots}{x^{5}}=\lim_{x
ightarrow0}(\frac{128}{15}-\frac{1024x^{2}}{315}+\cdots)$

As $x
ightarrow0$, the terms with $x$ (such as $-\frac{1024x^{2}}{315}+\cdots$) go to $0$

Answer:

$\frac{128}{15}$