<s>[INST] Find the limits (using Squeeze theorem or another theorem) I am not sure if I calculate the limits right. I would be grateful if somebody could check it.
Find the following limits:
1) $\lim_{x \to 0^{+}}\sqrt{x}cos(\frac{1}{x^2})$
Let us notice that $-\sqrt{x}<\sqrt{x}cos(\frac{1}{x^2})<\sqrt{x}$ and since:
$\lim_{x \to 0^{+}}\sqrt{x}=\lim_{x \to 0^{+}}-\sqrt{x}=0$,
we can write that $\lim_{x \to 0^{+}}\sqrt{x}cos(\frac{1}{x^2})=0$.
Could I just write $cos(\frac{1}{x^2})$ is bounded and $\lim_{x \to 0^{+}}\sqrt{x}=0$ and from this fact conclude that the product of these functions goes to 0 (instead of what I wrote above)?
2) $\lim_{x \to -\infty}\frac{sin(x^2)}{x}$
Let us notice that $\lim_{x \to -\infty}\frac{1}{x}=0$ and $sin(x^2)$ is bounded, so $\lim_{x \to -\infty}\frac{sin(x^2)}{x}=0$.
3)  $\lim_{x \to +\infty}\frac{2x+sin(x^2)}{3x+cos(\sqrt{x})}$
Let us notice that $\frac{2x-1}{3x+1}<\frac{2x+sin(x^2)}{3x+cos(\sqrt{x})}<\frac{2x+1}{3x-1}$ and since:
$\lim_{x \to +\infty}\frac{2x-1}{3x+1}=\lim_{x \to +\infty}\frac{2x+1}{3x-1}=2/3$,
we can write that $\lim_{x \to +\infty}\frac{2x+sin(x^2)}{3x+cos(\sqrt{x})}=2/3$.
4) $\lim_{x \to 0^{+}}\frac{2+sin(\frac{1}{x})}{x^3}$
We can write that $\frac{2-1}{x^3}<\frac{2+sin(\frac{1}{x})}{x^3}$, and since
$\lim_{x \to 0^{+}}\frac{1}{x^3}=+\infty$ so
$\lim_{x \to 0^{+}}\frac{2+sin(\frac{1}{x})}{x^3}=+\infty$.
 [/INST] For the first limit, you´re argument is definitely right, cause you can easily use
theorem, that if you can find neighborhood of point $x_{0},$ such that function $g(x)$ is bounded in there, and $lim_{x \rightarrow x_{0}} f(x)=0,$ then $lim _{x \rightarrow x_{0}}f(x)g(x)=0.$
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