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

find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it.\n\n$$\\lim_{x \\to 0} \\frac{e^{x}-e^{-x}-2 x}{x-\\sin (x)}$$\n\nresources\n\n16. -/1 points\n\nfind the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.\n\n$$\\lim_{x \\to 0} \\frac{\\cos (x)-1+\\frac{1}{2} x^{2}}{3 x^{4}}$$

find the limit. use lhospitals rule if appropriate. if there is a more elementary method, consider using it.\n\n$$\\lim_{x \\to 0} \\frac{e^{x}-e^{-x}-2 x}{x-\\sin (x)}$$\n\nresources\n\n16. -/1 points\n\nfind the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.\n\n$$\\lim_{x \\to 0} \\frac{\\cos (x)-1+\\frac{1}{2} x^{2}}{3 x^{4}}$$

Answer

Explanation:

Step1: Check if L'Hospital's Rule is applicable

When (x = 0), the numerator (e^{x}-e^{-x}-2x=e^{0}-e^{0}-2\times0 = 0) and the denominator (x-\sin(x)=0-\sin(0)=0). So, (\lim_{x\rightarrow0}\frac{e^{x}-e^{-x}-2x}{x - \sin(x)}) is in the (\frac{0}{0}) form.

Step2: Apply L'Hospital's Rule once

Differentiate the numerator and denominator. The derivative of (y = e^{x}-e^{-x}-2x) is (y^\prime=e^{x}+e^{-x}-2), and the derivative of (y=x - \sin(x)) is (y^\prime=1-\cos(x)). So, (\lim_{x\rightarrow0}\frac{e^{x}+e^{-x}-2}{1-\cos(x)})

Step3: Check the form again

When (x = 0), the numerator (e^{0}+e^{0}-2=0) and the denominator (1-\cos(0)=0). So, it is still in the (\frac{0}{0}) form.

Step4: Apply L'Hospital's Rule again

Differentiate the new - numerator and new - denominator. The derivative of (y = e^{x}+e^{-x}-2) is (y^\prime=e^{x}-e^{-x}), and the derivative of (y=1-\cos(x)) is (y^\prime=\sin(x)). So, (\lim_{x\rightarrow0}\frac{e^{x}-e^{-x}}{\sin(x)})

Step5: Check the form again

When (x = 0), the numerator (e^{0}-e^{0}=0) and the denominator (\sin(0)=0). So, it is still in the (\frac{0}{0}) form.

Step6: Apply L'Hospital's Rule for the third time

Differentiate the new - numerator and new - denominator. The derivative of (y = e^{x}-e^{-x}) is (y^\prime=e^{x}+e^{-x}), and the derivative of (y=\sin(x)) is (y^\prime=\cos(x)). So, (\lim_{x\rightarrow0}\frac{e^{x}+e^{-x}}{\cos(x)})

Step7: Evaluate the limit

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

Answer:

(2)