question 1\nby graphing the function\nf(x)=\frac{(cos x - cos 2x)}{x^{2}}\nand zooming in toward the point…

Explanation:

Step1: Recall trigonometric identities

We know that $\cos A-\cos B=- 2\sin\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right)$. So, $\cos x-\cos2x=-2\sin\left(\frac{x + 2x}{2}\right)\sin\left(\frac{x-2x}{2}\right)=2\sin\left(\frac{3x}{2}\right)\sin\left(\frac{x}{2}\right)$. Then $f(x)=\frac{\cos x-\cos2x}{x^{2}}=\frac{2\sin\left(\frac{3x}{2}\right)\sin\left(\frac{x}{2}\right)}{x^{2}}$.

Step2: Use the limit property $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$

We can rewrite $f(x)$ as $f(x)=\frac{2\sin\left(\frac{3x}{2}\right)\sin\left(\frac{x}{2}\right)}{x^{2}}=\frac{2\sin\left(\frac{3x}{2}\right)}{ \frac{3x}{2}}\cdot\frac{3}{2}\cdot\frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\cdot\frac{1}{2}$.

Step3: Calculate the limit as $x\rightarrow0$

$\lim_{x\rightarrow0}f(x)=\lim_{x\rightarrow0}\left(\frac{2\sin\left(\frac{3x}{2}\right)}{\frac{3x}{2}}\cdot\frac{3}{2}\cdot\frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}\cdot\frac{1}{2}\right)$. Since $\lim_{u\rightarrow0}\frac{\sin u}{u} = 1$, when $x\rightarrow0$, $\frac{3x}{2}\rightarrow0$ and $\frac{x}{2}\rightarrow0$. So $\lim_{x\rightarrow0}f(x)=\frac{3}{2}\times1\times1\times\frac{1}{2}\times2=\frac{3}{2}$.

Answer:

$\frac{3}{2}$