Chapter 14.2, Problem 56AYU

In problems 53-56, use the properties of limits and the facts that $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1\underset{x\to 0}{\lim }\frac{\cos x-1}{x}=0\underset{x\to 0}{\lim }\sin x=0\underset{x\to 0}{\lim }\cos x=1$ where $x$ is in radians, to find each limit.

$\underset{x\to 0}{\lim }\frac{{\sin }^{2}x+\sin x\left(\cos x-1\right)}{x^2}$

To determine

To find: The limit ${\lim }_{x\to \text{ 0}}\frac{{\sin }^{2}x+\sin x\left(\cos x–1\right)}{x^2}$, use the properties of limits and facts that ${\lim }_{x\to \text{ 0}}\frac{\sin x}{x}=1$; ${\lim }_{x\to \text{ 0}}\frac{\text{cos }x - 1}{x}=0$; ${\lim }_{x\to \text{ 0}}\sin x=0$; ${\lim }_{x\to \text{ 0}}\cos x=1$.

Answer to Problem 56AYU

Solution:

1

Explanation of Solution

Given:

$\underset{x\to 0}{\lim }\frac{{\sin }^{2}x+\sin x\left(\cos x–1\right)}{x^2}$

Formula used:

${\lim }_{x\to c}\left[f\left(x\right).g\left(x\right)\right]={\lim }_{x\to c}f\left(x\right).{\lim }_{x\to c}g\left(x\right)$

${\lim }_{x\to c}\left[f\left(x\right)+g\left(x\right)\right]={\lim }_{x\to c}f\left(x\right)+{\lim }_{x\to c}g\left(x\right)$

Calculation:

${\lim }_{x\to \text{ 0}}\frac{{\sin }^{2}x+\sin x\left(\cos x –1\right)}{x^2}$,

Now we have to simplify it,

$\frac{{\sin }^{2}x+\sin x\left(\cos x–1\right)}{x^2}=\frac{\sin x(\sin x+\left(\cos x–1\right))}{x^2}$

$=(\frac{\sin x}{x})(\frac{\text{sin }x+(\cos x-1)}{x})$

${\lim }_{x\to \text{ 0}}\frac{{\sin }^{2}x+\sin x\left(\cos x–1\right)}{x^2}={\lim }_{x\to \text{ 0}}[\frac{\sin x}{x}.\frac{\sin x+(\cos x-1)}{x}]$

$={\lim }_{x\to \text{ 0}}\frac{\sin x}{x}.{\lim }_{x\to \text{ 0}}\frac{\text{sin }x+ (\cos x-1)}{x}$

$={\lim }_{x\to \text{ 0}}\frac{\sin x}{x}.[{\lim }_{x\to \text{ 0}}\frac{\sin x}{x}+{\lim }_{x\to \text{ 0}}\frac{\cos x-1}{x}]$

$=1\left(1+0\right)=1$

${\lim }_{x\to \text{ 1}}\frac{{\sin }^{2}x+\sin x\left(\cos x –1\right)}{x^2}=1$