Results 1 to 5 of 5

Thread: [SOLVED] A D.N.E limit proofs

  1. #1
    Member
    Joined
    Mar 2008
    From
    Acolman, Mexico
    Posts
    118

    [SOLVED] A D.N.E limit proofs

    Hello,

    I need help to write a formal proof to show that the following limits do not exist.

    1. $\displaystyle \lim \limits_{x \rightarrow 0} \frac{1}{x}$. i.e. show that $\displaystyle \lim \limits_{x \rightarrow 0} \frac{1}{x}=L$ is false for $\displaystyle \forall L$.

    2. $\displaystyle \lim \limits_{x \rightarrow 1} \frac{1}{x-1}$ does not exist. i.e. show that $\displaystyle \lim \limits_{x \rightarrow 1} \frac{1}{x-1}$ is false for $\displaystyle \forall L$.

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1
    Consider the right-hand limit $\displaystyle \lim _{x \to 1^ + } \frac{1}{{x - 1}}$.
    The points $\displaystyle \left( {\forall n} \right)\left[ {x = 1 + \frac{1}{n}} \right]$ each can be used in that limit. But $\displaystyle \frac{1}{{1 + \frac{1}{n} - 1}} = n$.
    Thus that limit will get greater and greater. So the limit cannot exist.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,410
    Thanks
    1
    $\displaystyle \forall \epsilon > 0$, $\displaystyle \exists \delta > 0$ such that whenever $\displaystyle 0 < |x - 0| < \delta$ it necessarily follows that $\displaystyle \left|\frac{1}{x} - L\right| < e$

    $\displaystyle \left|\frac{1}{x} - L\right| < \epsilon$
    $\displaystyle L - \epsilon < \frac{1}{x} < L + \epsilon$
    $\displaystyle \frac{1}{L+\epsilon} < x < \frac{1}{L-\epsilon}$

    Let $\displaystyle \delta = \frac{1}{L - \epsilon}$

    So, we must find an $\displaystyle \epsilon$ such that the above will not be satisfied for any $\displaystyle \delta > 0$. Let $\displaystyle \epsilon = |L| + 1> 0$.

    Looking at L > 0
    $\displaystyle \delta = \frac{1}{L - (|L| + 1)} = \frac{1}{L - (L + 1)} = \frac{1}{-1} = -1 < 0$. Not possible.

    Or looking at L < 0
    $\displaystyle \delta = \frac{1}{L - (|L| + 1)} = \frac{1}{L - (-L + 1)} = \frac{1}{2L - 1} < 0$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    Let's do #2:

    Let's assume there is a number L such that

    $\displaystyle \lim_{x\rightarrow{1}}\frac{1}{x-1}=L$.

    Then there exists a number $\displaystyle \delta>0$ such that

    $\displaystyle |\frac{1}{x-1}-L|<1$ whenever $\displaystyle 0<|x-1|<{\delta}$

    In particular, $\displaystyle x=1+\frac{\delta}{\delta+1}$ and

    $\displaystyle x=1-\frac{\delta}{\delta+1}$ are two values of x.

    Therefore, $\displaystyle |\frac{\delta+1}{\delta}-L|<1$ and

    $\displaystyle |-\frac{\delta+1}{\delta}-L|<1$ or

    $\displaystyle \frac{1}{\delta}<L<\frac{1}{\delta}+2$ and

    $\displaystyle -2-\frac{1}{\delta}<L<\frac{-1}{\delta}$,

    which is impossible. Therefore, the limit does not exist.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2008
    From
    Acolman, Mexico
    Posts
    118
    Thank you very much for helping me out
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit proofs
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Feb 24th 2010, 03:26 PM
  2. limit identity proofs
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Dec 9th 2008, 12:06 PM
  3. Limit Proofs (Sup and Inf)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 19th 2008, 12:58 PM
  4. [SOLVED] two (tough?) limit proofs
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 13th 2008, 01:11 PM
  5. Real analysis - limit proofs
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 20th 2008, 06:29 PM

Search Tags


/mathhelpforum @mathhelpforum