1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

2 finding limits problems

  1. Jul 9, 2008 #1
    1. The problem statement, all variables and given/known data

    1. Find the limit of [tex]\lim_{x\rightarrow 0} \frac{1}{xe^{\frac{1}{x}}}[/tex]
    2. " " " " [tex]\lim_{x\rightarrow\infty} \frac {x}{\log_e x}[/tex]

    2. Relevant equations

    [tex]\lim_{x\rightarrow\infty} \frac{N}{x} = 0[/tex]

    [tex]\lim_{x\rightarrow n} x+a = \lim_{x\rightarrow n} x + \lim_{x\rightarrow n} a[/tex] etc

    3. The attempt at a solution

    1. I put in values of x close to 0, and as I approached from above I got values very close to 0, but when I approached from below the numbers became massively large and negative ([tex]f(-0.1)=-220264, f(-0.01)=-2.688\times10^{45}[/tex]). The answer in my book is zero, but my numbers say there is no limit as values of x approaching 0 do not approach the same number. Have I missed something out or is the book wrong?

    2. In the book the answer is "no limit", but I can't think of a way to evaluate it to prove it. The only thing I've thought of is dividing by x, but that did nothing and ended up going in circles :/
  2. jcsd
  3. Jul 9, 2008 #2


    User Avatar
    Homework Helper

    Have you learnt L Hospital's rule yet? Use it for both. Note that you have to express 1. in the correct form before you can use it.
  4. Jul 9, 2008 #3
    No, we don't learn that this year. This is last year high school stuff, and I was sick when the class was taught it so i'm trying to get through it myself. The notes up to this exercise in the book simply goes over what a limit is, evaluating by algebraic manipulation or solve for values of x and draw a graph/table, the equations listed above and cases of being careful with moduli.
  5. Jul 9, 2008 #4


    User Avatar
    Homework Helper

    Ok then perhaps we can take this somewhat intuitively. Consider the first question. As x->0, analyse the term in the denominator xe^(1/x). x will approach 0 and e^(1/x) will approach infinity, right? So we have two limits going in the opposite directions (very roughly speaking). But which one of these 2 would "reach its limit" faster? Which term would dominate?

    Alternatively, think of 1. as [tex]\frac{1/x}{e^{1/x}}[/tex]. Draw a graph of the numerator and that of the denominator on the same sketch. Which one would dominate as x->0?

    The second one you can also think of it intuitively. Look at the graph of y=x and y=ln x. What happens when x->infinity? Which one diverges faster?
  6. Jul 10, 2008 #5
    Ah I get it now, thanks!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?