find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it…

Explanation:

Step1: Check the form

When $x\rightarrow0$, the numerator $e^{x}-e^{-x}-2x=e^{0}-e^{-0}-2\times0 = 1 - 1-0=0$, and the denominator $x-\sin x=0-\sin(0)=0$. So, it is in the $\frac{0}{0}$ - form, and L'Hopital's Rule can be applied.

Step2: Differentiate numerator and denominator

Differentiate the numerator $y_1 = e^{x}-e^{-x}-2x$. Using the derivative rules $\frac{d}{dx}(e^{x})=e^{x}$, $\frac{d}{dx}(e^{-x})=-e^{-x}$ and $\frac{d}{dx}(ax)=a$, we get $y_1^\prime=e^{x}+e^{-x}-2$. Differentiate the denominator $y_2=x - \sin x$. Using the derivative rules $\frac{d}{dx}(x)=1$ and $\frac{d}{dx}(\sin x)=\cos x$, we get $y_2^\prime=1-\cos x$. So, the limit becomes $\lim_{x\rightarrow0}\frac{e^{x}+e^{-x}-2}{1 - \cos x}$.

Step3: Check the new - form and apply L'Hopital's Rule again

When $x\rightarrow0$, the numerator $e^{x}+e^{-x}-2=e^{0}+e^{-0}-2=1 + 1-2 = 0$, and the denominator $1-\cos x=1-\cos(0)=1 - 1=0$. So, it is still in the $\frac{0}{0}$ - form. Differentiate the new numerator $y_3=e^{x}+e^{-x}-2$. We get $y_3^\prime=e^{x}-e^{-x}$. Differentiate the new denominator $y_4=1-\cos x$. We get $y_4^\prime=\sin x$. So, the limit becomes $\lim_{x\rightarrow0}\frac{e^{x}-e^{-x}}{\sin x}$.

Step4: Check the new - form and apply L'Hopital's Rule again

When $x\rightarrow0$, the numerator $e^{x}-e^{-x}=e^{0}-e^{-0}=1 - 1=0$, and the denominator $\sin x=\sin(0)=0$. So, it is still in the $\frac{0}{0}$ - form. Differentiate the new numerator $y_5=e^{x}-e^{-x}$. We get $y_5^\prime=e^{x}+e^{-x}$. Differentiate the new denominator $y_6=\sin x$. We get $y_6^\prime=\cos x$. So, the limit becomes $\lim_{x\rightarrow0}\frac{e^{x}+e^{-x}}{\cos x}$.

Step5: Evaluate the limit

Substitute $x = 0$ into $\frac{e^{x}+e^{-x}}{\cos x}$. We have $\frac{e^{0}+e^{-0}}{\cos(0)}=\frac{1 + 1}{1}=2$.

Answer:

$2$