Question

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

$$\lim_{x \to 0} \frac{e^{x}-e^{-x}-2 x}{x-\sin (x)}$$

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1. -/1 points

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

$$\lim_{x \to 0} \frac{\cos (x)-1+\frac{1}{2} x^{2}}{3 x^{4}}$$

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, \(\lim_{x
ightarrow0}\frac{e^{x}-e^{-x}-2x}{x - \sin(x)}\) 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, \(\lim_{x
ightarrow0}\frac{e^{x}+e^{-x}-2}{1-\cos(x)}\)

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, \(\lim_{x
ightarrow0}\frac{e^{x}-e^{-x}}{\sin(x)}\)

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, \(\lim_{x
ightarrow0}\frac{e^{x}+e^{-x}}{\cos(x)}\)

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:

\(2\)