Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\frac{\cos x-1+\frac{1}{2}x^2}{x^4}=\frac{1}{24}$$
$$A=\lim_{x\to0}\frac{\cos x-1+\frac{1}{2}x^2}{x^4}$$ Here $\lim_{x\to0}(\cos x-1+\frac{1}{2}x^2)=\cos0-1+\frac{1}{2}0^2=1-1+0=0$ and $\lim_{x\to0}(x^4)=0^4=0.$ This limit is in an indeterminate form of $\frac{0}{0}$, a situation which can be helped by the use of L'Hospital's Rule: $$A=\lim_{x\to0}\frac{-\sin x-0+\frac{1}{2}\times2x}{4x^3}$$ $$A=\lim_{x\to0}\frac{-\sin x+x}{4x^3}$$ $\lim_{x\to0}(\sin x+x)=\sin0+0=0$ and $\lim_{x\to0}(4x^3)=4\times0^3=0.$ Again we have an indeterminate form of $\frac{0}{0}$. We continue to use L'Hospital's Rule: $$A=\lim_{x\to0}\frac{-\cos x+1}{12x^2}$$ $\lim_{x\to0}(-\cos x+1)=-\cos0+1=-1+1=0$ and $\lim_{x\to0}(12x^2)=12\times0^2=0.$ Indeterminate form of $\frac{0}{0}$. We use L'Hospital's Rule: $$A=\lim_{x\to0}\frac{-(-\sin x)+0}{24x}$$ $$A=\lim_{x\to0}\frac{\sin x}{24x}$$ $\lim_{x\to0}\sin x=\sin 0=0$ and $\lim_{x\to0}(24x)=24\times0=0.$ Sometimes it just goes on and on like this. Indeterminate form of $\frac{0}{0}$ for another application of L'Hospital's Rule: $$A=\lim_{x\to0}\frac{\cos x}{24}$$ $$A=\frac{\cos0}{24}$$ $$A=\frac{1}{24}$$