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

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\rightarrow0}\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\rightarrow0}\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\rightarrow0}\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\rightarrow0}\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})