# 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

$\lim_{x \to 0^{+}}\frac{2+sin(\frac{1}{x})}{x^3}=+\infty$.

• Yes, this is all correct. Well done. – Cornman Nov 2 '17 at 14:07
• I think your argument is good. – Eclipse Sun Nov 2 '17 at 14:08
• Thank you very much. And what with 1)? Should I use the Squeeze theorem or can I use the thoerem about a product of a bounded function and a convergent function? – SigmaMat Nov 2 '17 at 14:12

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 $lim _{x \rightarrow x_{0}}f(x)g(x)=0.$
• Sure, just for the squeeze, change $-\sqrt{x}<\sqrt{x}cos(\frac{1}{x^2})<\sqrt{x}$ to $-\sqrt{x}\leq \sqrt{x}cos(\frac{1}{x^2})\leq \sqrt{x}$. – stanly Nov 2 '17 at 14:29