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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?)

Michael Hardy
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dyrAnd
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    You can address this question ? where is this ? in which book ? (I think your work is correct ) – Khosrotash Aug 03 '15 at 17:21
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    The answer is $\frac{1}{2}$. Next time you can check the result by your self using wolfram:http://www.wolframalpha.com/ – idm Aug 03 '15 at 17:21
  • I think that it is originally from link. This is a set of tasks for first year undergraduate students in mathematics from the University of Warsaw in Poland. – dyrAnd Aug 03 '15 at 17:32
  • One small error, a typo since what follows is ok: all following sin in (*) should be in parentheses. – MasB Feb 25 '16 at 19:01

2 Answers2

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You did nothing wrong, so you are correct. Good job! The answer is $\frac 12$.

mathlove
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It would be absurd not to agree with everything said, but this is how I would approach this task:

$\begin{aligned}\lim\limits_{x\to\infty}\left(x\cdot\sin\left(\sqrt{x^2+3}-\sqrt{x^2+2}\right)\right)&=\lim\limits_{x\to\infty}\frac{\sin\left(x\left(\sqrt{1+\frac3{x^2}}-\sqrt{1+\frac2{x^2}}\right)\right)}{\frac1x}\\&=\lim\limits_{x\to\infty}\frac{\sin\left(\frac{\frac1{x}}{\sqrt{1+\frac3{x^2}}+\sqrt{1+\frac2{x^2}}}\right)}{\frac1{x}}\\&=\lim\limits_{x\to\infty}\frac{\sin\left(\frac1{x\left(\sqrt{1+\frac3{x^2}}+\sqrt{1+\frac2{x^2}}\right)}\right)}{\frac1{x\left(\sqrt{1+\frac3{x^2}}+\sqrt{1+\frac2{x^2}}\right)}}\cdot\frac1{\sqrt{1+\frac3{x^2}}+\sqrt{1+\frac2{x^2}}}\\&=\frac12\end{aligned}$ Which is essentially the same thing you did.

PinkyWay
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