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I'm having a really tough time trying to evaluate the limit for this expression

$\lim_{x\to\infty}\frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}$

The only hint I was given is that $\lim_{x\to\infty}\frac{\sin(x)}{x} = 0$.

By using trigonometric identities I can see that this leads to $\lim_{x\to\infty}\frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{\sin^2(\frac{1}{x})}$. I proceeded then to do multiply everything by $\frac{1}{x}$ which leaves me with $\lim_{x\to\infty}\frac{x^{-1}\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1}$ (according to the hint I was given).

However, now I don't know how to simplify the expression in the numerator. Any tips would be really appreciated.

Thanks

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  • $\begingroup$ The Simplest approach would be to get rid of $\infty$ by putting $t = 1/x$ so that $t \to 0+$ as $x \to \infty$. Then you need to do some simplification (as explained in various answers below) and use the following standard limit: $\lim_{t \to 0}(\sin t)/t = 1$ $\endgroup$ – Paramanand Singh Oct 5 '13 at 8:32
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Hint: In the argument of sine, multiply by $$\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}.$$

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$$\sin^2(\sqrt{x+1}-\sqrt{x})\sim \sin^2(1/(2\sqrt{x})) \sim {1\over4x},$$ as $x\to\infty$.

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$\sin^2\left(\dfrac{1}{x}\right) \approx \dfrac{1}{x^2}$ for very large $x$ rather than your $\dfrac{x^{-1}}{1}$ or perhaps $\dfrac{1}{x^{-1}}$.

A more helpful hint might have been: $\displaystyle\lim_{y\to 0}\dfrac{\sin(y)}{y} = 1$

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