find the limit as x approaches infinity f(x) = (e^-x)/x
\[f(x) = \frac{ e ^{-x} }{ x }\] No idea x_x
greaaaat :(
Haha, I hate functions also. I might find it in my book, let's see....
What does e^-x look like as x goes to infinity?
exactly! it's part of our calc homework for the prerequisites and I'm honestly clueless
my calculator says it's unable to graph e^-x..
what does e^-x represent what is another way to write this?
I have functions hwk also :/
could it be -xln? I'm so confused
i understand. its ok
I'm clueless also although I'm intrigued...
e^-x=1/(e^x)
oh! so you can treat x as a regular negative exponent.. ok, I'm following, now what?
so, do you see that as x gets larger and increases without bound, so too does the denominator.
still there?
still here, if the denominator get's bigger, than the answer get's smaller yes?
you got it!
now remember you have product of x and e^x in the denominator, so it will get smaller real real fast.
notice the abundant use of technical terms here. lol
so the original x in the denominator stays there so it's 1/(x)(e^x)
exactly; so if you have a fraction that has a numerator of 1 and a denominator or zillions, what is the limit of this quotient?
0?
see? you are very smart. you got the answer yourself. 0 is correct. Nicely done. Anything else i can help you with
the back of the book says it's 1 though....
either the problem is incorrect or the answer is incorrect.
are you familiar with L'Hopitals rule?
it says to use tables and graphs to find the answers.. find the limit as x approaches infinity when f(x) = e^-x/x then find the limit as x approaches f(x) from the left, then find any horizontal asymptotes if they exsist..... fml
never heard of it, why?
ok. we just found the horizontal asymptote
no you were right! i was looking at the wrong answers I think.. so confused
if you want a vertical asymptote, then you have to look at the quotient around x=0
it is ok. now, realize that you did most of this problem yourself. i kept nudging you in the right direction. have some faith in your own abilities
can i do anything else for you? Further explanation? any other questions?
thank you so much! hold on one moment,trying to record the work for this one, and then analyze the rest of the question
take your time. let me know if you have any questions as you record.
so that makes sense, but what about when it's approaching negative infinity? how could I figure that out?
ok. what does e^-x look like as x->\[-\infty\]?
do you see that the exponent increases positively without bound?
so it's infinity? but what about the x in the denominator? do I just ignore it?
no, never ignore it. But, we know that exponential functions increase much more rapidly than linear functions. Do you see that?
so why are we not including it? and only looking at e^-x?
of course, they increase exponentially ;)
because the exponential function gets so large so fast; think about it and take e^200 and divide by -200.. oops. from the left it will turn that quotient negative though, so the graph decreases without boutnd
bound*
does this make sense?
so multiplying by that extra x is just making it decrease even faster?
no, but because it divides a huge pos number, it just barely slows it down
you are on the right track here.
remember that the exponent is the neg of a negative
Plugging the equation into my calculator is just confusing me even more though, but that's how our teacher want's us to be able to prove it by graph.... uh.... head hurts
what does the graph look like on your calculator?
I think you only have to graph the function for large numbers to solve the above problem, dont u?
|dw:1347403916546:dw|
Join our real-time social learning platform and learn together with your friends!