Starting Limits HW Supposedly Easy

[Phyzwizz]

Just started learning limits today but I got stuck on this last problem...

Let f be the function given by f(x)=lnx/x for all x>0

Find lim f(x) as x-->0⁺

So I graphed the equation and plugged in points approaching zero on the right side but it seems to go on for infinity. Is the limit infinity? is that possible?

 

[LCKurtz]

Yes. In fact it goes to -∞.

 

[HallsofIvy]

Uh, no, the limit is neither [itex]\infty[/itex] nor [itex]-\infty[/itex]. The limit, as x goes to 0, of ln(x) is [itex]-\infty[/itex] (which may be what LCKurtz meant) but that x in the denominator makes the limit 0. x goes to [itex]\infty[/itex] "faster" than ln(x) goes to [itex]-\infty[/itex].

 

[Phyzwizz]

Awesome, thanks for the help guys. That clears things up. Oh and this doesn't have anything to do with limits, but how do you do the custom equations thing that makes the math equations on this site look so clean and wonderful.

 

[gb7nash]

Uh, no, the limit is neither [itex]\infty[/itex] nor [itex]-\infty[/itex]. The limit, as x goes to 0, of ln(x) is [itex]-\infty[/itex] (which may be what LCKurtz meant) but that x in the denominator makes the limit 0. x goes to [itex]\infty[/itex] "faster" than ln(x) goes to [itex]-\infty[/itex].

I don't follow this. If we're talking about:

[tex]\frac{ln(x)}{x}[/tex]

As x -> 0⁺, the numerator approaches -inf and the denominator approaches +0, so you should obtain -inf?

 

[LCKurtz]

Uh, no, the limit is neither [itex]\infty[/itex] nor [itex]-\infty[/itex]. The limit, as x goes to 0, of ln(x) is [itex]-\infty[/itex] (which may be what LCKurtz meant) but that x in the denominator makes the limit 0. x goes to [itex]\infty[/itex] "faster" than ln(x) goes to [itex]-\infty[/itex].

The limit is -∞. x → 0 in the denominator giving a -∞/0⁺ form.