Formal limits (1 Viewer)

untouchablecuz · Mar 8, 2010

$\\$Define $f:\mathbb{R}\to \mathbb{R},f(x)=\frac{\sin 3x}{x^2+4}\\$Prove that $\lim_{x\to \infty }f(x)=0 $ using the formal definition of a limit.$\\\\\\$Suppose $\epsilon >0 $ and $x>0,\\|f(x)-L|=\left|\frac{\sin 3x}{x^2+4}\right|\leq \left|\frac{3}{x^2+4}\right|<\frac{3}{x^2}<\epsilon \Rightarrow x>\sqrt{\frac{3}{\epsilon }}\\$If $M=\sqrt{\frac{3}{\epsilon }},\\$Then $|f(x)-L|<\epsilon $ for $x>M\\\\$Q.E.D.$$

is this correct in both terms of what i'm doing and its layout?

Pwnage101 · Mar 8, 2010

$\sin \left (3x \right )\leq 1\;\; \forall x\in \mathbb{R}$

but yes.

Affinity · Mar 19, 2010

Replace L by 0 and.. the implication has to go the other way
remember it's for all epsilon's there's a M such that |f(x) - limit| < epsilon for all x > M

Formal limits (1 Viewer)

untouchablecuz

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Pwnage101

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Affinity

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