Limit?!

**dhiab** · January 4th 2010, 10:04 PM

Calculate this limit :

$\mathop {\lim }\limits_{x \to + \infty } \left( {\sin \left( {\pi \sqrt {x^2 + x} } \right)} \right)$

**tonio** · January 4th 2010, 11:19 PM

Originally Posted by dhiab

Calculate this limit :

$\mathop {\lim }\limits_{x \to + \infty } \left( {\sin \left( {\pi \sqrt {x^2 + x} } \right)} \right)$

The limit doesn't exist: take the sequence $\{x_n\}=\left\{\,\frac{-1+\sqrt{4n^2+1}}{2}\,\right\}$ $\Longrightarrow \sqrt{x_n^2+x_n}=n\Longrightarrow \sin(\pi\sqrt{x_n^2+x_n})=\sin n\pi=0$ , and now

take the sequence $\{y_n\}=\left\{\frac{-1+\sqrt{n^2+1}}{2}\right\}\Longrightarrow \sqrt{y_n^2+y_n}=\frac{n}{2}$ $\Longrightarrow \sin(\pi\sqrt{y_n^2+y_n})=\sin\frac{n}{2}\pi=\pm 1$

Tonio

Ps. In fact it is enough to take the sequence $\{y_n\}$ ...

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