## anonymous one year ago a question about limit.

1. anonymous

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

2. anonymous

@ganeshie8 .

3. Astrophysics

is that a floor function or just being multiplied

4. anonymous

multiplied

5. Astrophysics

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

6. Astrophysics

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

7. anonymous

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

8. Astrophysics

What?

9. Astrophysics

So that is a floor function..

10. Astrophysics

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

11. ganeshie8

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

12. ganeshie8

Incase if you want to find the limit $$x\to\infty$$, similar reasoning yields : $\lim_{x \rightarrow -\infty} (2x-1)\left[ \frac{ -5 }{ x-2 } \right] = \lim_{x \rightarrow \infty} (2x-1)*(-1) = -\infty$