Help finding the limit of trigonometric $(\cos(7x))^\frac{1}{x^2}$ without l'Hospital

Question

I don't know how to address it without l'Hospital, currently I don't know how to approach this problem, to what form I need to transform it to. $$ \lim_{x\to 0}\left(\cos(7x)\right)^{\frac{1}{x^2}} $$

Answer 1

We have $\cos (7x)\to 1$ and $1/x^2\to \infty$ as $x\to 0$, so $$\lim_{x\to 0}(\cos 7x)^{1/x^2}=e^{\lambda}\text{ with }\lambda=\lim_{x\to 0}\frac{(\cos 7x) -1}{x^2}$$ Now, use $1-\cos \epsilon\underbrace{\sim}_{\epsilon \to 0} \epsilon^2/2$.

Answer 2

HINT:

Set $7x=2h$

$$(\cos7x)^{1/x^2}=\left((\cos2h)^{h^2}\right)^{49/4}$$

$\displaystyle(\cos2h)^{h^2}=\left((1-2\sin^2h)^{-\frac1{2\sin^2h}}\right)^{-2\left(\frac{\sin h}h\right)^2}$

Now use $\lim_{h\to0}(1+h)^{1/h}=e$

Answer 3

hint: first taking $\log $ we have: $f(x) = \dfrac{\log(\cos (7x))}{x^2}= \dfrac{\log(1-2\sin^2(\frac{7x}{2}))}{-2\sin^2(\frac{7x}{2})}\cdot \dfrac{-2\sin^2(\frac{7x}{2})}{x^2}$. Can you see some well-known formula poping up here?

Answer 4

Just another way to do it.

Considering $$A=\left((\cos(kx)\right)^{\frac 1 {x^2}}\implies \log(A)={\frac 1 {x^2}}\log(\cos(kx))$$ Now consider Taylor expansion $$\cos(t)=1-\frac{t^2}{2}+\frac{t^4}{24}+O\left(t^6\right)$$ $$\cos(kx)=1-\frac{k^2 x^2}{2}+\frac{k^4 x^4}{24}+O\left(x^6\right)$$ Now, using Taylor again $$\log(\cos(kx))=-\frac{k^2 x^2}{2}-\frac{k^4 x^4}{12}+O\left(x^6\right)$$ $$\log(A)=-\frac{k^2}{2}-\frac{k^4 x^2}{12}+O\left(x^4\right)$$ Finally, using Taylor again $$A=e^{\log(A)}=e^{-\frac{k^2}{2}}-\frac{1}{12} \left(e^{-\frac{k^2}{2}} k^4\right) x^2+O\left(x^4\right)$$ which shows the limit and how it is approached.