Recent content by coverticus

hey thanks for all your help guys, i got the answer (e^-1) finally and it was for a quiz so i appreciate it.

ok I got that much, and I got x/(1/ln(x/(x+1))), but I'm lost as to evaluate that lim with l'Hospitals, would it be (1) / (1/ln(x/(x+1)))', if so I haven't the slightest as to how to get that derivative.

so you would get e^limx->inf of xln(x/(x+1)) and then what?

I should have posted this sooner, but isn't it in the form 1^inf ?

Homework Statement Evaluate lim x\rightarrow infinity of (\frac{x}{x+1})^{}x, state explicitly the type of the indeterminate form. Homework Equations The Attempt at a Solution I somewhat understand how to use l'Hospital's rule when the form is 0/0, but the inf/inf throws me off...

what I think I have it. ln(e^-x) = -x, so -x = ln(1/2), or x = -ln(1/2). But what about the zero?

ok so you take ln(e^-x) = ln(1/2)? What about the other critical number, you can't take ln(0) so how do you solve that function? Or is it a critical number because it's not defined? This is very confusing to me.

ok, so after substituting y for e^-x i saw that f has two critical points, on at x=0, and the other where e^-x=(1/2) . But how do you figure out x?

so you can factor out the e^-x and your left with: e^-x(2(1)^2-1) = 0, correct?

ok thanks, if you feel so inclined you could help me with another question I have on here. https://www.physicsforums.com/showthread.php?t=194855

Homework Statement Find the absolute maximum and minimum values of f(x) = e^{}-x - e^{}-2x on [0,1]. f '(x) = -e^-x + 2e^-2x Homework Equations The Attempt at a Solution f '(x) = -e^-x + 2e^-2x 0 = -e^-x + 2e^-2x e^-x = 2e^-2x ln(e^-x) = ln(2e^-2x) -x = ?

how can you do that without creating an extrema?

does that satisfy the continuity?

I still need somewhat of a solid answer here, do I just sketch a graph where x=0 on [1,5] or something different? Any help here would be great.

Yes it is [1,5], and x^3 has a critical point at 0, and it has a slope of zero. Correct?