Answer
The value of the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}$ is $\frac{2}{3}$.
Work Step by Step
Consider the limit to be solved,
$\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}$
Simplify it as follows:
$\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x}+\frac{1}{3}\times \frac{\cos x-1}{x} \right)$
Using $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$
$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x}+\frac{1}{3}\times \frac{\cos x-1}{x} \right) \\
& =\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x} \right)+\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{3}\times \frac{\cos x-1}{x} \right)
\end{align}$
Using $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)\times g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\times \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$
$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x}+\frac{1}{3}\times \frac{\cos x-1}{x} \right) \\
& =\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{2}{3}\times \frac{\sin x}{x} \right)+\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{3}\times \frac{\cos x-1}{x} \right) \\
& =\underset{x\to 0}{\mathop{\lim }}\,\frac{2}{3}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}+\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{3}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x-1}{x}
\end{align}$
Using the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}=1\ \text{and }\underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x-1}{x}=0$, also using limit property $\underset{x\to a}{\mathop{\lim }}\,c=c,c\text{ is constant}$.
$\begin{align}
& \underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2}{3}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{x}+\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{3}\times \underset{x\to 0}{\mathop{\lim }}\,\frac{\cos x-1}{x} \\
& =\frac{2}{3}\times 1+\frac{1}{3}\times 0 \\
& =\frac{2}{3}+0 \\
& =\frac{2}{3}
\end{align}$
Thus, the value of the limit $\underset{x\to 0}{\mathop{\lim }}\,\frac{2\sin x+\cos x-1}{3x}$ is $\frac{2}{3}$.
