Limit Problem: (1 - cos t)/(t^2) as t goes to 0

Hello, sallyk57!

Evaluate: \(\displaystyle \,\lim_{x\to0}\frac{1\,-\,cos x}{x^2}\)
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Multiply top and bottom by \(\displaystyle (1\,+\,\cos x):\)

\(\displaystyle \L\;\;\frac{1\,-\,\cos x}{x^2}\,\cdot\,\frac{1\,+\,\cos x}{1\,+\,\cos x}\;=\;\frac{\1\,-\,\cos^2x}{x^2(1\,+\,\cos x)}\;=\;\frac{\sin^2x}{x^2(1\,+\,\cos x)} \;=\;\left(\frac{\sin x}{x}\right)^2\frac{1}{1\,+\,\cos x}\)

Therefore: \(\displaystyle \L\,\lim_{x\to0}\left[\left(\frac{\sin x}{x}\right)^2\frac{1}{1\,+\,\cos x}\right] \;= \;(1)^2\cdot\frac{1}{1\,+\,1}\;=\;\frac{1}{2}\)
 
Here's another slightly different approach using L'Hopital.

\(\displaystyle \L\\\lim_{x\to\0}\frac{1-cos(x)}{x^{2}}\)

Take derivative of numerator and denominator.

\(\displaystyle \L\\(1-cos(x))dx=sin(x)\)

\(\displaystyle \L\\x^{2}dx=2x\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{1}{2}\frac{sin(x)}{x}\)

\(\displaystyle \L\\\frac{1}{2}\lim_{x\to\0}\frac{sin(x)}{x}\)

We could stop here because of the familiar limit(as Soroban showed), but I will trudge onward.

Now, do L'Hopital again by taking derivatives of numerator and denominator:

\(\displaystyle \L\\sin(x)dx=cos(x)\) and \(\displaystyle xdx=1\)

\(\displaystyle \L\\\frac{1}{2}\lim_{x\to\0}(cos(x))\)

\(\displaystyle \H\\\frac{1}{2}\)
 
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