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Limits of a function

  1. May 19, 2015 #1
    1. The problem statement, all variables and given/known data
    1. f(x) = (1+x)1/x
    lim x->0+ (1+x)1/x

    2. lim x->0+ 10+ln(x)

    2. Relevant equations


    3. The attempt at a solution
    1. lim x->0+ (1+x)1/x

    (1+0+)1/0+

    I really dont understand how the final answer is e... thanks for helping.

    = e1


    2. lim x->0+ 10+ln(x)

    ln(0+) = -infinite
    10 - infinite = - infinite
    I dont understand why ln(x) is - infinite is there any rule for that or thats something I have to know by heart ?
     
  2. jcsd
  3. May 19, 2015 #2

    Svein

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    1. If you let [itex] u=\frac{1}{x}[/itex], then [itex]\lim_{x\rightarrow 0+} (1+x)^{\frac{1}{x}} [/itex] transforms into [itex]\lim_{u\rightarrow +\infty}(1+\frac{1}{u})^{u} [/itex]. Does this expression look familiar? (http://en.wikipedia.org/wiki/E_(mathematical_constant)).
    2. Do the same transformation here ( [itex] u=\frac{1}{x}[/itex]). The expression changes to [itex]\lim_{u\rightarrow +\infty}(10+ln(\frac{1}{u}))=10+\lim_{u\rightarrow +\infty}ln(\frac{1}{u})=10-\lim_{u\rightarrow +\infty}ln(u) [/itex].
     
  4. May 19, 2015 #3

    SammyS

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    For item #2.

    I don't know whether or not you want to consider it as " knowing by heart " , but if you're serious about math, physics, engineering, etc., you should be very familiar with the characteristics of logarithmic functions.

    ln(x) is strictly increasing, it's domain is the interval ##\displaystyle \ (0,\ \infty) \ ##, it's range is ##\displaystyle \ (-\infty,\ \infty) \ ##.

    That's enough to get ##\displaystyle \ \lim_{x\to0^+} \ln(x)=-\infty \ ##.

    I or someone else will cover item #1 in a separate post.
     
  5. May 19, 2015 #4
    ##x = e^{\ln x}##. In our current problem, ##x## represents the expression your are taking the limit of.
    In the process ##x\to 0## , the expression ##\ln (1+x)## is equivalent to ##x##.

    E: This is basic mathematical analysis and not about physics. Not being mean, just pointing it out.

    Also, usually ##x= e^{\ln |x|}##, however here we approach ##0## strictly from the right side hence the ##x## is positive regardless.
     
    Last edited: May 19, 2015
  6. May 19, 2015 #5

    vela

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    You should be able to reason it out. The relation ##y = \ln x## is equivalent to ##x = e^y##. If you want to make x very very small, i.e. close to 0, what kind of value do you have to use for y?
     
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