$\lim_{\rightarrow \infty} \sin(x²)$
1 Answer
The following limit
\[\lim_{x\rightarrow \infty}\sin x^2\]
does not exists as $\sin (x^2)$ oscillates between $-1$ and $1$ as $x \rightarrow \infty.$ However,
\[0=\lim_{x\rightarrow \infty}-\frac{1}{x} \leq \lim_{x\rightarrow \infty}\frac{\sin x^2}{x} \leq \lim_{x\rightarrow \infty}\frac{1}{x}=0.\]
Hence
\[\lim_{x\rightarrow \infty}\frac{\sin x^2}{x}=0.\]
Savionf
574
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thanks a lot
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