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Lim (x/x+1)^x

  1. Oct 16, 2009 #1
    Can someone help me with finding the limit of (x/x+1)^x as x tends to infinity
     
  2. jcsd
  3. Oct 16, 2009 #2

    Mark44

    Staff: Mentor

    The limit is infinity, unless of course this is what you meant:
    [tex]\lim_{x \rightarrow \infty} \left ( \frac{x}{x + 1}\right )^x [/tex]

    If you want to write the quotient of x and x + 1, put parentheses around what goes in the denominator, like so: x/(x + 1).
     
  4. Oct 16, 2009 #3
    The reciprocal is
    [tex]\lim_{x \rightarrow \infty} \left ( \frac{x+1}{x}\right )^x [/tex]
    Can you do that limit?
     
  5. Oct 16, 2009 #4

    Yes, i meant (x/(x+1))^x
     
    Last edited: Oct 16, 2009
  6. Oct 16, 2009 #5
    Never mind, I solved it.

    (x/(x(1+1/x)))^x = (1/(1+1/x))^x = 1^x/(1+1/x)^x where 1^x = 1 and (1+(1/x))^x as x tends to inf = e^1

    therefore as x tends to infinity the function tends to e^-1
     
  7. Oct 18, 2009 #6
    [tex]\lim_{x\rightarrow \infty} (x/x+1)^x= \lim _{x\rightarrow \infty} (1+\frac{-1}{x+1})^{-(x+1)\frac{x}{-(x+1)}})=\lim_{x\rightarrow \infty} e^\frac{x}{-(x+1)}}=e^{-1} [/tex]
     
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