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Thread: Limit

  1. #1
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    Limit

    Prove that if $\displaystyle f(x)<g(x)<h(x)$ and $\displaystyle \lim_{x \to \infty}f(x) = \lim_{x \to \infty}h(x)=L$, then $\displaystyle \lim_{x \to \infty}g(x)=L$.

    Can we prove this by contradiction, assuming to the contrary that $\displaystyle \lim_{x \to \infty}g(x)\not = L$?
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  2. #2
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    Here are several observations.
    We have $\displaystyle 0 \leqslant g(x) - f(x)\, \leqslant h(x) - f(x) \Rightarrow \left| {g(x) - f(x)} \right| \leqslant \left| {h(x) - f(x)} \right|\,$.

    We know that if $\displaystyle \varepsilon > 0$ then $\displaystyle \left( {\exists K > 0} \right)\left[ {x \geqslant K\, \Rightarrow \,\left| {f(x) - L}\right| < \frac{\varepsilon }{3}\;\& \,\left| {h(x) - L} \right| < \frac{\varepsilon }{3}} \right]$.

    Note that $\displaystyle \left| {g(x) - L} \right| \leqslant \left| {g(x) - f(x)} \right| + \left| {f(x) - L} \right| \leqslant \left| {h(x) - f(x)} \right| + \left| {f(x) - L} \right|$.

    Can you finish?
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  3. #3
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    You did 99% of it. The rest is easy.

    I did it by proving $\displaystyle \lim_{x \to \infty}(h(x)-f(x) )= 0$, then I showed that

    $\displaystyle |h(x)-g(x)-(L-L)| =|h(x)-g(x)|\leq |h(x)|-|g(x)|<|h(x)-f(x)-(L-L)|=|h(x)-f(x)| < ?$

    Then realized that there wasn't direct connection to $\displaystyle \epsilon$--hmm not too smart.
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  4. #4
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    $\displaystyle \dfrac{\varepsilon }
    {3} + \dfrac{\varepsilon }
    {3} + \dfrac{\varepsilon }
    {3} = \varepsilon $
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