# Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$

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.

\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}

• You use the assumption that d.d.d. x $sinx = x$, I understand correctly? Dec 31 '18 at 14:02
• 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. Dec 31 '18 at 14:07

Letting $$h=\frac1x$$:

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

• Actually the searched limit is equal to $$\frac{1}{2}$$ Dec 31 '18 at 13:36
• Fixed.${ }$${ }$ Dec 31 '18 at 13:37

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 $$\frac{1}{2}+ O(\frac{1}{x})$$.

\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}