Find the limit of the function without L'Hôpital's rule

Question

I have a problem, such as:

$$\lim_{x \to 0} \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}$$

How do I solve this problem without using L'Hôpital's rule or small-o?

Thanks!

Answer 1

HINT

We can use that

$$\left(\frac{\cos(ax)}{\cos(bx)}\right) ^\tfrac{1}{x^2}=\left(\frac{1-(1-\cos(ax))}{1-(1-\cos(bx))}\right) ^\tfrac{1}{x^2}$$

and then use that

$$1-(1-\cos(ax))^\tfrac{1}{x^2}=\left(1-(1-\cos(ax))^\tfrac1{1-\cos(ax)}\right)^{a^2\tfrac{1-\cos(ax)}{(ax)^2}}$$

$$1-(1-\cos(bx))^\tfrac{1}{x^2}=\left(1-(1-\cos(bx))^\tfrac1{1-\cos(bx)}\right)^{b^2\tfrac{1-\cos(bx)}{(bx)^2}}$$

and refer to standard limits.

Answer 2

In terms of $t:=\lim_{y\to0}\frac{\ln\cos y}{y^2}$, write your limit as $L=\exp\ell$ with$$\ell:=\lim_{x\to0}\frac{\ln\tfrac{\cos ax}{\cos bx}}{x^2}=t(a^2-b^2).$$We now prove $t=-\tfrac12$ so $L=\exp\tfrac{b^2-a^2}{2}$. In terms of $z:=1-\cos y$,$$t=\underbrace{\lim_{z\to0}\frac{\ln(1-z)}{-z}}_{1}\underbrace{\lim_{y\to0}\frac{\cos y-1}{y^2}}_{-\tfrac12}.$$

Answer 3

Basically a variant of user's answer: Using familiar limits $(1-u)^{1/u}\to e^{-1}$ and ${\sin u\over u}\to1$ as $u\to0$, we have

$$\begin{align} (\cos ax)^{1/x^2} &=(\cos^2ax)^{1/(2x^2)}\\ &=(1-\sin^2ax)^{1/(2x^2)}\\ &=\left((1-\sin^2ax)^{1/\sin^2ax}\right)^{(a^2/2)(\sin ax/(ax))^2}\\ &\to (e^{-1})^{(a^2/2)\cdot1^2}\\ &=e^{-a^2/2} \end{align}$$

So in general we have

$$\left(\cos ax\over\cos bx\right)^{1/x^2}={(\cos ax)^{1/x^2}\over(\cos bx)^{1/x^2}}\to{e^{-a^2/2}\over e^{-b^2/2}}=e^{(b^2-a^2)/2}$$

Answer 4

Composing Taylor series $$y= \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}\implies\log(y)=\frac{1}{x^2}\big[\log(\cos(ax))-\log(\cos(bx))\big]$$

Using $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\cos(cx)=1-\frac{c^2 x^2}{2}+\frac{c^4 x^4}{24}+O\left(x^6\right)$$ $$\log(\cos(ax))=-\frac{c^2 x^2}{2}-\frac{c^4 x^4}{12}+O\left(x^6\right)$$

Use it twice (for $c=a$ and $c=b$ and subtract to get $$\log(\cos(ax))-\log(\cos(bx))=\frac{1}{2} x^2 \left(b^2-a^2\right)+\frac{1}{12} x^4 \left(b^4-a^4\right)+O\left(x^6\right)$$ $$\log(y)=\frac{1}{2} \left(b^2-a^2\right)+\frac{1}{12} x^2 \left(b^4-a^4\right)+O\left(x^4\right)$$ $$y=e^{\log(y)}=e^{\frac {b^2-a^2}2}\left(1+\frac{1}{12} x^2 \left(b^4-a^4\right)+O\left(x^4\right)\right)$$ which shows the limit and also how it is approached.

Answer 5

$$\lim_{x\to0}\cos(ax)^{1/x^2}=\lim_{t\to0}\cos(t)^{a^2/t^2}=\lim_{t\to0}(\cos(t)^{1/t^2})^{a^2}=\left(\lim_{t\to0}(\cos(t)^{1/t^2}\right)^{a^2}$$ and the answer is of the form

$$c^{a^2-b^2}.$$

We can find $c$ with

$$(\cos(t))^{1/t^2}=\left(\left(1-2\sin^2\left(\frac t2\right)\right)^{1/\sin^2(\frac t2)}\right)^{\sin^2(\frac t2)/4(\frac t2)^2}\to (e^{-2})^{1/4}.$$