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I know that $$\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})\\=\lim\limits_{ x\to + \infty}x\cdot \sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right).$$ If $x \rightarrow + \infty$, then $\sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right)\rightarrow \sin0 $. However I have also $x$ before $\sin x$ and I don't know how to calculate it.

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\begin{align} \lim_{x \to \infty} x \cdot \left( \sin \left( \sqrt{x^2+3}-\sqrt{x^2+2}\right)\right)&=\lim_{x \to \infty} x \cdot \sin \left( \frac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)\\ &=\lim_{x \to \infty} \frac{x}{\sqrt{x^2+3}+\sqrt{x^2+2}}\\ &=\lim_{x \to \infty} \frac{1}{\sqrt{1+\frac{3}{x^2}}+\sqrt{1+\frac{2}{x^2}}}\\ &= \frac12 \end{align}

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  • $\begingroup$ You use the assumption that d.d.d. x $sinx = x$, I understand correctly? $\endgroup$
    – MP3129
    Dec 31 '18 at 14:02
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    $\begingroup$ not sure what do those $d$ mean.The main trick is $\lim_{h \to 0^+} \frac{\sin h}h=1$, hence that is why I can remove the sine. $\endgroup$ Dec 31 '18 at 14:07
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Letting $h=\frac1x$:

$$\begin{array}{cl} &\displaystyle \lim_{x \to \infty} x \sin \left( \sqrt{x^2+3} - \sqrt{x^2+2} \right) \\ =&\displaystyle \lim_{x \to \infty} x \sin \left( \frac 1 {\sqrt{x^2+3} + \sqrt{x^2+2} } \right) \\ =&\displaystyle \lim_{h \to 0^+} \frac1h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\ =&\displaystyle \lim_{h \to 0^+} \frac 1 {\sqrt{1+3h^2} + \sqrt{1+2h^2}} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\ =&\displaystyle \frac12 \times \lim_{h \to 0^+} \frac {\sqrt{1+3h^2} + \sqrt{1+2h^2}} h \sin \left( \frac h {\sqrt{1+3h^2} + \sqrt{1+2h^2} } \right) \\ =&\displaystyle \frac12 \times 1 \\ =&\dfrac12 \end{array}$$

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  • $\begingroup$ Actually the searched limit is equal to $$\frac{1}{2}$$ $\endgroup$ Dec 31 '18 at 13:36
  • $\begingroup$ Fixed.${ }$${ }$ $\endgroup$
    – Kenny Lau
    Dec 31 '18 at 13:37
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The ordo calculus is very useful for such problems. Sometimes you do not need the full power of Taylor series, only the first couple of coefficients.

$\sqrt{x^2+3}- \sqrt{x^2+2} = x(\sqrt{1+3/x^2}- \sqrt{1+2/x^2})=$

$= x(1+\frac{1}{2}\cdot(3/x^2)+O(1/x^4)- 1-\frac{1}{2}\cdot(2/x^2)+O(1/x^4))= \frac{1}{2x}+O(1/x^3)$.

Now $\sin(\frac{1}{2x}+O(1/x^3))= \frac{1}{2x}+O(1/x^2)$ as $x\rightarrow \infty$, thus the expression is $\frac{1}{2}+ O(\frac{1}{x})$.

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\begin{align} x \sin \left( \sqrt{x^2+3}-\sqrt{x^2+2}\right) &=x \sin \left( \frac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)\\ &=\left(\dfrac{x}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right) \dfrac{\sin \left( \dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)} {\left(\dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}\\ &=\left(\dfrac{1}{\sqrt{1+\frac{3}{x^2}}+\sqrt{1+\frac{2}{x^2}}}\right) \dfrac{\sin \left( \dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)} {\left(\dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}\\ &\to \dfrac 12 \cdot 1 \ \text{as $x \to \infty$}\\ &\to \dfrac 12 \ \text{as $x \to \infty$} \end{align}

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