Limit Problem 1


Compute \underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1-\cos x} ? 😀

Solution

Since {{\sin }^{2}}x=1-{{\cos }^{2}}x

We multiply the numerator and denominator by 1+\cos x and use the fact that \underset{x\to 0}{\mathop{\lim }}\,\frac{x}{\sin x}=1 Therefore:

\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}(1+\cos x)}{(1-\cos x)(1+\cos x)}=\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{x}{\sin x} \right)}^{2}}\underset{x\to 0}{\mathop{\lim }}\,(1+\cos x)=2

**Another solution, using L’Hopital’s rule, is possible: \underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{2}}}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2x}{\sin x}=2

Thank you 😀

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About Bunchhieng

I like math, programming and web design!!
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