Question

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it.
\\( \lim _ { x \
ightarrow 0 } \frac { \cos ( x ) - 1 + \frac { 1 } { 2 } x ^ { 2 } } { 3 x ^ { 4 } } \\)

Explanation:

Step1: Check the form of the limit

Substitute \(x = 0\) into \(\frac{\cos(x)-1+\frac{1}{2}x^{2}}{3x^{4}}\). We get \(\frac{\cos(0)-1+\frac{1}{2}(0)^{2}}{3(0)^{4}}=\frac{1 - 1+0}{0}=\frac{0}{0}\), so L'Hospital's Rule can be applied.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and the denominator.
The derivative of \(y=\cos(x)-1+\frac{1}{2}x^{2}\) is \(y'=-\sin(x)+x\).
The derivative of \(y = 3x^{4}\) is \(y'=12x^{3}\).
So the limit becomes \(\lim_{x
ightarrow0}\frac{-\sin(x)+x}{12x^{3}}\). Substitute \(x = 0\), we get \(\frac{-\sin(0)+0}{12(0)^{3}}=\frac{0}{0}\), apply L'Hospital's Rule again.

Step3: Apply L'Hospital's Rule the second - time

Differentiate the numerator and the denominator.
The derivative of \(y=-\sin(x)+x\) is \(y'=-\cos(x)+1\).
The derivative of \(y = 12x^{3}\) is \(y'=36x^{2}\).
So the limit becomes \(\lim_{x
ightarrow0}\frac{-\cos(x)+1}{36x^{2}}\). Substitute \(x = 0\), we get \(\frac{-\cos(0)+1}{36(0)^{2}}=\frac{0}{0}\), apply L'Hospital's Rule again.

Step4: Apply L'Hospital's Rule the third - time

Differentiate the numerator and the denominator.
The derivative of \(y=-\cos(x)+1\) is \(y'=\sin(x)\).
The derivative of \(y = 36x^{2}\) is \(y'=72x\).
So the limit becomes \(\lim_{x
ightarrow0}\frac{\sin(x)}{72x}\). Substitute \(x = 0\), we get \(\frac{\sin(0)}{72(0)}=\frac{0}{0}\), apply L'Hospital's Rule again.

Step5: Apply L'Hospital's Rule the fourth - time

Differentiate the numerator and the denominator.
The derivative of \(y=\sin(x)\) is \(y'=\cos(x)\).
The derivative of \(y = 72x\) is \(y'=72\).
So the limit becomes \(\lim_{x
ightarrow0}\frac{\cos(x)}{72}\).

Step6: Evaluate the limit

Substitute \(x = 0\) into \(\frac{\cos(x)}{72}\), we have \(\frac{\cos(0)}{72}=\frac{1}{72}\).

Answer:

\(\frac{1}{72}\)