Any idea how to find $\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$?

Question

$$\lim_{x\to 0} \frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}$$

I am trying to solve this limit for 2 days, but still cant find the solution which is $\sqrt{2}$ (that's what is written in the solution sheet)

I tried multiplying with the conjugate, tried with some identities but nothing much because of that $x^2$ in the $\cos$. Then i tried L'Hopital because it is $\frac{0}{0}$ and still that $\cos$ in the square root is doing problems. I tried on symbolab calculator but it can't solve it.

So can someone help me how to solve this? Thank you.

Answer 1

if you multiply and divide by both conjugates you get:

$$\lim_{x \to 0} \frac{(1+\cos{x})\sin{(x^2)}}{\sqrt{1+\cos{(x^2)}} \space \sin^2{x}}$$

Amplifying to get limit of the form $\frac{\sin{x}}{x} :$

$$=\lim_{x \to 0} \frac{(1+\cos{x})}{\sqrt{1+\cos{x^2}}} \space \frac{\sin{(x^2)}}{x^2} \space \frac{x^2}{\sin^2{x}}=\frac{2}{\sqrt{2}}*1*1^2=\sqrt{2}$$

Answer 2

An approach based on the fact that $$ \lim_{u\to0} \frac{1-\cos u}{u^2} = \frac{1}{2} \tag{1} $$ which is a standard limit (and equivalent to a Taylor expansion of $\cos$ to order $2$ at $0$).

$$ \frac{\sqrt{1-\cos(x^2)}}{1-\cos x} = \sqrt{\frac{1-\cos(x^2)}{x^4}}\cdot\frac{x^2}{1-\cos x} \xrightarrow[x\to0]{}\sqrt{\frac{1}{2}}\cdot\frac{1}{\frac{1}{2}} = \sqrt{2}\,, $$ applying (1) to $u=x^2$ and $u=x$ separately (for the first, since $x^2\to 0$ when $x\to 0$).

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To be completely precise: we used (1) twice, and the continuity of both $\sqrt{\cdot}$ and the inverse function at $1/2$ to have $$\lim_{x\to 0} \sqrt{g(x)}\frac{1}{h(x)} = \lim_{x\to 0}\sqrt{g(x)}\lim_{x\to 0}\frac{1}{h(x)} = \sqrt{\lim_{x\to 0}g(x)}\frac{1}{\lim_{x\to 0}h(x)}$$

Answer 3

By L'Hopital's rule, we have $$\begin{aligned} \lim_{x \to 0}\frac{1 - \cos(x^2)}{(1 - \cos(x))^2} &= \lim_{x \to 0}\frac{2x\sin(x^2)}{2(1 - \cos(x))\sin(x)} \\ &= \lim_{x \to 0}\frac{x}{\sin(x)}\frac{\sin(x^2)}{1 - \cos(x)} \\ &= \lim_{x \to 0}\frac{\sin(x^2)}{1 - \cos(x)} \\ \end{aligned}$$ where we have used that $$\lim_{x \to 0}\frac{x}{\sin(x)} = 1$$ Now we can apply L'Hopital's rule again to obtain $$\begin{aligned} \lim_{x \to 0}\frac{\sin(x^2)}{1 - \cos(x)} &= \lim_{x \to 0} \frac{2x\cos(x^2)}{\sin(x)} \\ &= \lim_{x \to 0} \frac{x}{\sin(x)} 2\cos(x^2) \\ &= \lim_{x \to 0} 2\cos(x^2) \\ &= 2 \end{aligned}$$ Summarizing what we have so far, $$\lim_{x \to 0} \frac{1 - \cos(x^2)}{(1 - \cos(x))^2} = 2$$ Now take the square root of both sides and use the fact that $\sqrt{(\cdot)}$ is continuous at zero (note that we approach only from the right) to conclude that $$ \begin{aligned} \lim_{x \to 0}\frac{\sqrt{1 - \cos(x^2)}}{1 - \cos(x)} &= \lim_{x \to 0}\sqrt{\frac{1 - \cos(x^2)}{(1 - \cos(x))^2}} \\ &= \sqrt{\lim_{x \to 0} \frac{1 - \cos(x^2)}{(1 - \cos(x))^2} } \\ &= \sqrt{2} \end{aligned}$$ as desired.

Answer 4

\begin{align*} \lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)} &=\lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}\cdot(1+\cos x)}{1-\cos^2(x)} =2\lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}}{\sin^2(x)}\\[10pt] &=2\lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}}{x^2}\cdot\frac{x^2}{\sin^2(x)} =2\lim_{x\to 0}\frac{\sqrt{1-\cos(x^2)}}{x^2} \stackrel{y:=x^2}{=}2\lim_{y\to 0^+}\frac{\sqrt{1-\cos(y)}}{y}\\[10pt] &=2\lim_{y\to 0^+}\sqrt{\frac{1-\cos(y)}2}\cdot\frac{\sqrt 2}{y} =2\lim_{y\to 0^2}\sin\left(\frac y2\right)\cdot\frac{\sqrt 2}{y} =\sqrt 2\lim_{y\to 0^+}\sin\left(\frac y2\right)\cdot\frac{2}{y}=\boldsymbol{\sqrt 2} \end{align*}

Answer 5

Just note that $1-\cos t=2\sin^2(t/2)$ and hence the expression under limit is equal to $$\frac{\sqrt{2}\sin(x^2/2)}{2\sin^2(x/2)}$$ which can be rewritten as $$\frac{1}{\sqrt{2}}\cdot\frac{\sin(x^2/2)}{x^2/2}\cdot \frac {x^2}{2}\cdot\frac{4}{x^2}\cdot\frac{(x/2)^2}{\sin^2(x/2)}$$ Using limit $\lim_{t\to 0}(\sin t) /t=1$ we get the desired limit as $$\frac{1}{\sqrt{2}}\cdot 1\cdot 2\cdot 1^2=\sqrt{2}$$

Answer 6

If you want to go beyond the limit it self, use Taylor series and binomial expansion $$\cos(x^2)=1-\frac{x^4}{2}+\frac{x^8}{24}+O\left(x^{12}\right)$$ $$1-\cos(x^2)=\frac{x^4}{2}-\frac{x^8}{24}+O\left(x^{12}\right)$$ $$\sqrt{1-\cos \left(x^2\right)}=\frac{x^2}{\sqrt{2}}-\frac{x^6}{24 \sqrt{2}}+O\left(x^{10}\right)$$ $$1-\cos(x)=\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)$$ $$\frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}=\frac{\frac{x^2}{\sqrt{2}}-\frac{x^6}{24 \sqrt{2}}+O\left(x^{10}\right) }{ \frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)}$$ Now, long division to get $$\frac{\sqrt{1-\cos(x^2)}}{1-\cos(x)}=\sqrt{2}+\frac{x^2}{6 \sqrt{2}}+O\left(x^4\right)$$ which shows the limit and how it is approached.

For the fun, use your pocket calculator for $x=\frac \pi 6$. The exact value would be $$2 \sqrt{2} \left(2+\sqrt{3}\right) \sin \left(\frac{\pi ^2}{72}\right)\approx 1.44244$$ while the above truncated series would give $$\frac{432+\pi ^2}{216 \sqrt{2}}\approx 1.44652$$