Question

a question about limit.

Answer 1

\[\lim_{x \rightarrow -\infty} (2x-1)\left[ \frac{ -5 }{ x-2 } \right]\]

Answer 2

@ganeshie8 .

Answer 3

is that a floor function or just being multiplied

Answer 4

multiplied

Answer 5

Oh why not just multiply through and then expand \[\frac{ -5(2x-1) }{ x-2 } = \frac{ -10x }{ x-2 }+\frac{ 5 }{ x-2 }\] then you can take the limits separately I was being lazy by not putting limits you shouldn't do that

Answer 6

The first one -10x/(x-2) might have some trouble but you can let \[x-2 = x\left( 1-\frac{ 2 }{ x } \right)\] then you can cancel out the x's and then just plug in stuff with limits

Answer 7

no .-5/(x-2) is in brackets .that means [3.14]=3

Answer 8

What?

Answer 9

So that is a floor function..

Answer 10

Eh I have to go but with this problem @jtvatsim and @ganeshie8 can help you, they're great.

Answer 11

Notice that the function \(\left[\dfrac{-5}{x-2}\right] \) 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 \(0\) :

\[\lim_{x \rightarrow -\infty} (2x-1)\left[ \frac{ -5 }{ x-2 } \right] = \lim_{x \rightarrow -\infty} (2x-1)*0 = 0\]

Answer 12

Incase if you want to find the limit \(x\to\infty\), similar reasoning yields :

\[\lim_{x \rightarrow -\infty} (2x-1)\left[ \frac{ -5 }{ x-2 } \right] = \lim_{x \rightarrow \infty} (2x-1)*(-1) = -\infty\]