How can I solve $\lim_{x \to \infty} (\cos \frac {1}{x})^{x^2}$ without differentiation/l'Hospital/Taylor Series?

Question

The given limit is: $$\lim_{x \to \infty} \left (\cos \dfrac {1}{x} \right)^{\displaystyle x^2} $$

The answer is $\dfrac{1}{\sqrt e}.$

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My working: $$\lim_{x \to \infty} \left (\cos \dfrac {1}{x} \right)^{\displaystyle x^2} $$ $$=\lim_{x \to \infty} e^{\displaystyle \ln{\left(\cos \dfrac {1}{x} \right)^{\displaystyle x^2}}} $$ $$=\lim_{x \to \infty} e^{\displaystyle \ln{\left( \sqrt {1-\sin^2 \dfrac {1}{x}} \right)^{\displaystyle x^2}}} $$ $$=\lim_{x \to \infty} e^{\displaystyle \ln{\left({1-\sin^2 \dfrac {1}{x}} \right)^{\dfrac {x^2}{2}}}} $$

I am stuck at the 4th limit and I don't know how to go on. What should I do?

Answer 1

I would write $$\left[\left(\cos\left(\frac{1}{x}\right)^{1/x}\right)\right]^{x^3}$$

Answer 2

Do the substitution $x=1/t$, so the limit of the logarithm becomes $$ \lim_{t\to0^+}\frac{\log\cos t}{t^2}=\frac{1}{2}\lim_{t\to0}\frac{\log(1-\sin^2t)}{t^2}=\frac{1}{2}\lim_{t\to0^+}\frac{\log(1-\sin^2t)}{\sin^2t}\frac{\sin^2t}{t^2} $$ Now you can use $$ \lim_{u\to0}\frac{\log(1-u)}{u}=-1 $$