<s>[INST] limit $x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2})$ as $x\rightarrow \infty $ I want to calculate limit $x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2})$ as $x \rightarrow \infty $ without L'Hôpital's rule.
I found this task on the Internet. The answer given by the author is $2$.
$$\lim_{x\rightarrow\infty} x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2}) = \\
= \lim_{x\rightarrow\infty} x \cdot \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}}\cdot(\sqrt{x^2+3}-\sqrt{x^2+2}) =  (*)\\
$$
at this point I took care of a sine interior
$$ \lim_{x\rightarrow\infty} \sqrt{x^2+3}-\sqrt{x^2+2} = \\ 
=\lim_{x\rightarrow\infty} \frac{(\sqrt{x^2+3}-\sqrt{x^2+2}) \cdot(\sqrt{x^2+3} + \sqrt{x^2+2})}{(\sqrt{x^2+3}+\sqrt{x^2+2})} = \\ 
= \lim_{x\rightarrow\infty} \frac{1}{(\sqrt{x^2+3}+\sqrt{x^2+2})} = \left[\frac{1}{\infty}\right] = 0 \\ 
 $$
and whole sin function
$$ \lim_{x\rightarrow\infty} \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}} = \left[\frac{\sin(y\rightarrow0)}{y\rightarrow0}\right] = 1 $$
finally
$$(*) =\lim_{x\rightarrow\infty} \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}}\cdot  \frac{x}{(\sqrt{x^2+3}+\sqrt{x^2+2})} = \\ = \lim_{x\rightarrow\infty} \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}}\cdot  \frac{x}{x \left(\sqrt{1+\frac{3}{x^2}}+\sqrt{1+\frac{1}{x^2}}\right)} = \frac{1}{2}  $$
Unfortunately $\frac{1}{2} \neq 2$ so who is wrong? (and if I'm wrong, then why?)
 [/INST] You did nothing wrong, so you are correct. Good job! The answer is $\frac 12$.
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