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I want to evaluate the following oscillating limit x tends to infinity $$\sin(\sqrt{x+1})-\sin(\sqrt{x})$$ I tried evaluating this limit using trigonometric transformations but didn't arrive at the answer

Keith
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  • I have fixed your MathJAX in the most reasonable-looking way. Is the current expression the one you intended? – Patrick Stevens Mar 31 '16 at 13:35
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    I've tried to figure out what is being asked here from looking through the edits, and it doesn't make any sense. Adding sines as you did completely changes the question, so none of the answers already given apply any longer (and did you really want $\sin(\sqrt{x+1} - \sin\sqrt{x})$ and not $\sin(\sqrt{x+1} - \sqrt{x})$)? And what happened to the limit? you still talk about them, but you don't actually define a limit in your problem. What is it you are trying to do with this expression? – Paul Sinclair Mar 31 '16 at 15:13
  • I'm voting to close this question as off-topic because the poster changed completely the question after receiving answers. –  Mar 31 '16 at 15:49

3 Answers3

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$$\lim_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)=\lim_{x\to\infty}\left(\sqrt{x+1}-\sqrt{x}\right)\left(\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}\right)=$$ $$\lim_{x\to\infty}\frac{\left(\sqrt{x+1}-\sqrt{x}\right)\left(\sqrt{x+1}+\sqrt{x}\right)}{\sqrt{x+1}+\sqrt{x}}=$$ $$\lim_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=\lim_{t\to\infty}\frac{1}{\sqrt{t}+\sqrt{t}}=$$ $$\lim_{t\to\infty}\frac{1}{2\sqrt{t}}=\frac{1}{2}\lim_{t\to\infty}\frac{1}{\sqrt{t}}=\frac{1}{2}\cdot0=0$$

Jan Eerland
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Let $x=\cot^22y$ as $x\to\infty,y\to0^+$

$$\lim_{x\to\infty}(\sqrt{x+1}-\sqrt x)=\lim_{y\to0^+}(\csc2y-\cot2y)=\lim_{y\to0^+}\dfrac{2\sin^2y}{2\sin y\cos y}$$

Cancel out $\sin y$ safely as $y\to0,y\ne0,\sin y\to0,\sin y\ne0$

lab bhattacharjee
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$$\sqrt{x+1}-\sqrt{x}=\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x})}{\sqrt{x+1}+\sqrt{x}}=\frac{x+1-x}{\sqrt{x+1}+\sqrt{x}}=\frac{1}{\sqrt{x+1}+\sqrt{x}}$$

Jimmy R.
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