## Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to0}\frac{x^2}{1-\cos x}=2$$
$$A=\lim_{x\to0}\frac{x^2}{1-\cos x}$$ L'Hospital's Rule would be more effective here since it involves both $x$ and $\cos x$. Since $\lim_{x\to0}(x^2)=0^2=0$ and $\lim_{x\to0}(1-\cos x)=1-\cos0=1-1=0,$ this limit is an indeterminate form of $\frac{0}{0}$, so according to L'Hospital's Rule: $$A=\lim_{x\to0}\frac{\frac{d}{dx}(x^2)}{\frac{d}{dx}(1-\cos x)}$$ $$A=\lim_{x\to0}\frac{2x}{0-(-\sin x)}$$ $$A=\lim_{x\to0}\frac{2x}{\sin x}$$ Again, since $\lim_{x\to0}(2x)=2\times0=0$ and $\lim_{x\to0}(\sin x)=\sin 0=0$, we meet another indeterminate form of $\frac{0}{0}$, so we apply L'Hospital's Rule one more time: $$A=\lim_{x\to0}\frac{\frac{d}{dx}(2x)}{\frac{d}{dx}(\sin x)}$$ $$A=\lim_{x\to0}\frac{2}{\cos x}$$ $$A=\frac{2}{\cos 0}$$ $$A=\frac{2}{1}=2$$