Question

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations.

$\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{{\mathbf{sin}}^{\mathbf{2}}\mathbf{2}\mathbf{x}}{{\mathbf{x}}^{\mathbf{2}}}$ .

Short answer

The value of the limit is 4.

Step-by-step solution

Given Information

The function is $\frac{{\sin }^{2}2x}{x^2}$.

formula used

The formula is mentioned below as:

$\mathbf{sin}\left(x\right)\mathbf{=}\mathbf{x}\mathbf{-}\frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{6}}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{5}}}{\mathbf{120}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$

Find the value

The formula is mentioned below.

$\sin \left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+....$

The function becomes as follows.

$$\begin{gathered} {\sin }^2\left(x\right)=\left[x-\frac{x^3}{6}+\frac{x^5}{120}+...\right].\left[x-\frac{x^3}{6}+\frac{x^5}{120}+...\right] \\ {\sin }^2\left(x\right)=x^2-\frac{x^4}{3}+\frac{2x^6}{45}+... \\ {\sin }^2\left(2x\right)=4x^2-\frac{16x^4}{3}+\frac{64x^6}{45}+... \\ \frac{{\sin }^2\left(2x\right)}{x^2}=4-\frac{16x^2}{3}+\frac{64x^4}{45}+... \end{gathered}$$

Apply the limit in the above equation, and substitute the value of x. The limit becomes as follows.

$$\begin{gathered} \underset{x\to 0}{lim}\frac{{\sin }^2\left(2x\right)}{x^2}={\left(4-\frac{16x^2}{3}+\frac{64x^4}{45}+.....\right)}_{x-o} \\ =4 \end{gathered}$$

The value of the limit is 4 .