Question

by using squeeze theorem shaw that the lim of 1-coax/x² =0 as x tends to 0

Answer 1

$We\,\,have:\\t-\frac{t^3}{6}\leqslant \sin t\leqslant t,t\geqslant 0\\from\,\,which\\1-\frac{t^2}{6}\leqslant \frac{\sin t}{t}\leqslant 1\\Next,\\\frac{1-\cos x}{x^2}=\frac{2\sin ^2\frac{x}{2}}{x^2}=\frac{1}{2}\left( \frac{\sin \left| \frac{x}{2} \right|}{\left| \frac{x}{2} \right|} \right) ^2\\\frac{1}{2}\left( \frac{\sin \left| \frac{x}{2} \right|}{\left| \frac{x}{2} \right|} \right) ^2\leqslant \frac{1}{2}\\\frac{1}{2}\left( \frac{\sin \left| \frac{x}{2} \right|}{\left| \frac{x}{2} \right|} \right) ^2\geqslant \frac{1}{2}\left( 1-\frac{\left| \frac{x}{2} \right|^2}{6} \right) ^2=\frac{1}{2}\left( 1-\frac{x^2}{24} \right) ^2\\Since\\\underset{x\rightarrow 0}{\lim}\frac{1}{2}=\frac{1}{2},\underset{x\rightarrow 0}{\lim}\frac{1}{2}\left( 1-\frac{x^2}{24} \right) ^2=\frac{1}{2},\\by\,\,the\,\,squeeze\,\,theorem\\\underset{x\rightarrow 0}{\lim}\frac{1-\cos x}{x^2}=\frac{1}{2}\\$