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$\lim \limits_{x \to -\infty} \log(\cos\frac1x)(x^3-3x+\sin x)$

Is L'Hôpital's rule a way to evaluate this limit? Any suggestions would be appreciated.

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First note that $$\frac{x^3-3x+\sin x}{x^3}\rightarrow 1$$

so it is a case of evaluating

$$\lim\limits_{x\rightarrow -\infty}x^3\ln(\cos \frac{1}{x})$$

Let $y=\frac{1}{x}$ then we have $y\rightarrow 0^{-}$ and we have the function

$$\frac{\ln \cos y}{y^3}$$

Now $$\frac{\ln(1+\cos y -1)}{\cos y -1}\rightarrow 1$$

So we are left with $$\frac{\cos y -1}{y^3}$$ but since $$\frac{\cos y -1}{y^2}\rightarrow -\frac{1}{2}$$ this limit will be $+\infty$.

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