find the limit. use lhospitals rule where app...

# Explanation: ## Step1: Check the form of the limit When \(x\rightarrow0\), we substitute \(x = 0\) into \(\frac{e^{x}+e^{-x}-2}{e^{x}-x - 1}\). We get \(\frac{e^{0}+e^{-0}-2}{e^{0}-0 - 1}=\frac{1 + 1-2}{1-0 - 1}=\frac{0}{0}\), so L'Hopital's Rule can be applied. ## Step2: Differentiate the numerator and denominator Differentiate the numerator \(y_1=e^{x}+e^{-x}-2\), \(y_1^\prime=e^{x}-e^{-x}\). Differentiate the denominator \(y_2=e^{x}-x - 1\), \(y_2^\prime=e^{x}-1\). Now we find the limit \(\lim_{x\rightarrow0}\frac{e^{x}-e^{-x}}{e^{x}-1}\). ## Step3: Check the form of the new - limit Substitute \(x = 0\) into \(\frac{e^{x}-e^{-x}}{e^{x}-1}\), we get \(\frac{e^{0}-e^{-0}}{e^{0}-1}=\frac{1 - 1}{1-1}=\frac{0}{0}\), so we apply L'Hopital's Rule again. ## Step4: Differentiate the new numerator and denominator Differentiate the numerator \(y_3=e^{x}-e^{-x}\), \(y_3^\prime=e^{x}+e^{-x}\). Differentiate the denominator \(y_4=e^{x}-1\), \(y_4^\prime=e^{x}\). Now we find the limit \(\lim_{x\rightarrow0}\frac{e^{x}+e^{-x}}{e^{x}}\). ## Step5: Evaluate the limit Substitute \(x = 0\) into \(\frac{e^{x}+e^{-x}}{e^{x}}\), we have \(\frac{e^{0}+e^{-0}}{e^{0}}=\frac{1 + 1}{1}=2\). # Answer: 2