Evaluate:
$$\lim_{x\to 0}\dfrac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$$
I have tried to simplify the expression using the identity $1-\cos(x) = 2 \sin^2 (x/2)$, but I have still failed to remove the indeterminate form.
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$\begingroup$ Reading math.meta.stackexchange.com/questions/5020 is a good way of getting acquainted with Latex. $\endgroup$– Angina SengNov 16, 2018 at 17:57
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$\begingroup$ You may use D' Hospital $\endgroup$– dmtriNov 16, 2018 at 18:15
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$\begingroup$ *L'Hopitals rule $\endgroup$– Henry LeeNov 16, 2018 at 19:09
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4 Answers
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Hint: $$1-\cos\alpha=2\sin^2\dfrac{\alpha}{2}$$ Apply this with both numerator and denominator. Then use $$\lim_{x\to 0}\dfrac{\sin x}{x}=\lim_{x\to 0}\dfrac{x}{\sin x}=1$$
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hint
$$\lim_{X\to 0}\frac{1-\cos(X)}{X^2}=\frac 12$$
and $$\sqrt{1-\cos(x^2)}=x^2\sqrt{\frac{1-\cos(x^2)}{(x^2)^2}}$$
You will find $\sqrt{2}$.
answered Nov 16, 2018 at 18:05
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A little more details:
$$\lim_{x \to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)} = \lim_{x \to 0} \frac{\sqrt{2\sin^2(\frac{x^2}{2})}}{2 \sin^2 \frac{x}{2}} = \frac{\sqrt{2}}{2} \lim_{x \to 0} \frac{\sin(\frac{x^2}{2})}{\sin^2 \frac{x}{2}} = \sqrt{2}\lim_{x \to 0} (\frac{\sin(\frac{x^2}{2})}{\frac{x^2}{2}}\frac{\frac{x^2}{4}}{\sin^2\frac{x}{2}})$$
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Multiply by conjugates: $$\lim_{x\to 0}\dfrac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}=\lim_{x\to 0}\dfrac{\sin(x^2)}{\sin^2(x)}\cdot \frac{1+\cos(x)}{\sqrt{1+\cos(x^2)}}=\\ \lim_{x\to 0}\dfrac{\sin(x^2)}{x^2}\cdot \dfrac{x^2}{\sin^2(x)}\cdot\frac{2}{\sqrt{2}}=\sqrt{2}.$$
