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anonymous
 one year ago
a question about limit.
anonymous
 one year ago
a question about limit.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow \infty} (2x1)\left[ \frac{ 5 }{ x2 } \right]\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0is that a floor function or just being multiplied

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Oh why not just multiply through and then expand \[\frac{ 5(2x1) }{ x2 } = \frac{ 10x }{ x2 }+\frac{ 5 }{ x2 }\] then you can take the limits separately I was being lazy by not putting limits you shouldn't do that

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0The first one 10x/(x2) might have some trouble but you can let \[x2 = x\left( 1\frac{ 2 }{ x } \right)\] then you can cancel out the x's and then just plug in stuff with limits

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no .5/(x2) is in brackets .that means [3.14]=3

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0So that is a floor function..

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Eh I have to go but with this problem @jtvatsim and @ganeshie8 can help you, they're great.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Notice that the function \(\left[\dfrac{5}{x2}\right] \) is identically equal to \(0\) for \(x\lt 3\). Since this is just the definition of floor function, and our interest is in finding limit as \(x\to\infty\), we may replace that piece by \(0\) : \[\lim_{x \rightarrow \infty} (2x1)\left[ \frac{ 5 }{ x2 } \right] = \lim_{x \rightarrow \infty} (2x1)*0 = 0\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Incase if you want to find the limit \(x\to\infty\), similar reasoning yields : \[\lim_{x \rightarrow \infty} (2x1)\left[ \frac{ 5 }{ x2 } \right] = \lim_{x \rightarrow \infty} (2x1)*(1) = \infty\]
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