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Problem of the week #92 - December 31st, 2013

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Jameson

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Jan 26, 2012
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Thank you to anemone for this problem! :)

Calculate $\displaystyle \lim_{x \to \infty} (\sin \sqrt{x+1}-\sin \sqrt{x})$.
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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) magneto

Solution (from MarkFL):
Let:

\(\displaystyle L=\lim_{x\to\infty}\left(\sin\left(\sqrt{x+1} \right)-\sin\left(\sqrt{x} \right) \right)\)

Application of the sum-to-product identity:

\(\displaystyle \sin(\alpha)-\sin(\beta)=2\sin\left(\frac{\alpha-\beta}{2} \right)\cos\left(\frac{\alpha+\beta}{2} \right)\)

And the limit property:

\(\displaystyle \lim_{x\to c}\left(k\cdot f(x) \right)=k\cdot\lim_{x\to c}\left(f(x) \right)\)

Allows us to write:

\(\displaystyle L=2\lim_{x\to\infty}\left(\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2} \right)\cos\left(\frac{\sqrt{x+1}+\sqrt{x}}{2} \right) \right)\)

Rationalization of the numerator of the sine function gives us:

\(\displaystyle L=2\lim_{x\to\infty} \left( \sin \left(\frac{1}{2 \left( \sqrt{x+1}+ \sqrt{x} \right)} \right) \cos \left(\frac{ \sqrt{x+1}+ \sqrt{x}}{2} \right) \right)\)

Now, using the property of limits:

\(\displaystyle \lim_{x\to c}\left(f(x)\cdot g(x) \right)=\left(\lim_{x\to c}\left(f(x) \right) \right)\left(\lim_{x\to c}\left(g(x) \right) \right)\)

We obtain:

\(\displaystyle L=2 \lim_{x \to\infty} \left( \sin \left( \frac{1}{2 \left( \sqrt{x+1}+ \sqrt{x} \right)} \right) \right) \lim_{x \to\infty} \left( \cos \left( \frac{ \sqrt{x+1}+ \sqrt{x}}{2} \right) \right)\)

The first limit goes to zero, and the second limit is bounded, hence:

\(\displaystyle L=0\)

Alternate solution using calculus (from magneto):
Define $f(x) := \sin \sqrt{x}$. The limit can be rewritten as $f(x+1)-f(x)$. Since
$f$ is continuous and differentiable on $\mathbb{R}^+$, by the Mean Value Theorem, we can find
$c \in [x, x+1]$ where:
\begin{equation}
f'(c) = \frac{f(x+1)-f(x)}{(x+1)-x}.
\end{equation}
For any given $x$, we can find $c \geq x$, so the equation can be rewritten as:
\begin{equation}
\sin \sqrt{x+1} - \sin \sqrt{x} = f'(c) = \frac{\cos \sqrt{c}}{2 \sqrt{c}} \leq \frac{\cos\sqrt{c}}{2\sqrt{x}}.
\end{equation}
Therefore,
\begin{equation}
\bigl|{\sin \sqrt{x+1} - \sin \sqrt{x}} \bigr| \leq \frac{\bigl|{\cos\sqrt{c}}\bigr|}{\bigl|{2\sqrt{x}}\bigr|} \leq \frac{1}{2\sqrt{x}}.
\end{equation}
We know that $\lim_{x \to \infty} \frac{1}{2\sqrt{x}} = 0$, so by the Squeeze Theorem and last equation
we have that:
\begin{equation}
\lim_{x \to \infty} (\sin \sqrt{x+1} - \sin \sqrt{x}) = 0.
\end{equation}
 
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