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Determine the limit.
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\[\lim_{x \rightarrow \infty} {{2x^2 - 3} \over {4x^3 +5x}}\]
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\[\lim_{x\rightarrow\infty}\frac{2x^2-3}{4x^3+5x}=\lim_{x\rightarrow\infty}\frac{2-3/x^2}{4x+5/x}=\lim_{x\rightarrow\infty}\frac{2}{4x}=0\]
why \[/x^2\] and \[/x\] ?
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ahhh! I get it!
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