# Science:Math Exam Resources/Courses/MATH100/December 2016/Question 05 (c)/Solution 1

Using trigonometric identity , the given function can be written as

Since we have ,

by the continuity of the square function, we obtain the desired limit as

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Using trigonometric identity $1-\cos ^{2}x=\sin ^{2}x$, the given function can be written as

${\begin{aligned}{\frac {1-\cos ^{2}x}{x^{2}}}&={\frac {\sin ^{2}x}{x^{2}}}=\left({\frac {\sin x}{x}}\right)^{2}.\end{aligned}}$

Since we have $\lim _{x\to 0}{\frac {\sin x}{x}}=\lim _{x\to 0}{\frac {\sin x-\sin 0}{x-0}}=\left.{\frac {d}{dx}}(\sin x)\right|_{x=0}=\left.\cos x\right|_{x=0}=1$,

by the continuity of the square function, we obtain the desired limit as

${\begin{aligned}\lim _{x\to 0}{\frac {1-\cos ^{2}x}{x^{2}}}=\lim _{x\to 0}\left({\frac {\sin x}{x}}\right)^{2}=\left(\lim _{x\to 0}{\frac {\sin x}{x}}\right)^{2}=\color {blue}1.\end{aligned}}$

- This page was last edited on 19 October 2017, at 19:02.