Text Compute limx→01−cosxx2\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x^{2}}x→0limx21−cosx. Solution Multiply numerator and denominator by 1+cosx1+\cos x1+cosx to use the identity 1−cos2x=sin2x1-\cos^{2}x=\sin^{2}x1−cos2x=sin2x: 1−cosxx2=(1−cosx)(1+cosx)x2(1+cosx)=sin2xx2(1+cosx)=(sinxx)2⋅11+cosx.\frac{1-\cos x}{x^{2}}=\frac{(1-\cos x)(1+\cos x)}{x^{2}(1+\cos x)}=\frac{\sin^{2}x}{x^{2}(1+\cos x)}=\left(\frac{\sin x}{x}\right)^{2}\cdot\frac{1}{1+\cos x}.x21−cosx=x2(1+cosx)(1−cosx)(1+cosx)=x2(1+cosx)sin2x=(xsinx)2⋅1+cosx1. As x→0x\to 0x→0: (sinxx)2→1\left(\tfrac{\sin x}{x}\right)^{2}\to 1(xsinx)2→1 and 11+cosx→12\tfrac{1}{1+\cos x}\to\tfrac{1}{2}1+cosx1→21, so L=1⋅12=12.L=1\cdot\frac12=\frac12.L=1⋅21=21. L=12 \boxed{\,L=\tfrac12\,}L=21