I used u substitution for the limit lim and got the limit does not exist by saying u=\frac{1}{x}. Is this correct and if so would that mean \lim\limits_{x \to\infty}x^3 \sin\frac{1}{x^2} and \lim\limits_{x \to\infty}x^3 \sin\frac{1}{x} also don't exist?
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real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}
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