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Proving the limit laws

Addition. Let ER be a subset of the real numbers, f,g:ER be real functions, and L1,L2R be real numbers. Now suppose lim and \lim_{x\to a} g(x) = L_2. We would like to show that \lim_{x\to a} (f(x)+g(x)) = L_1+L_2.

Since \lim_{x\to a} f(x) = L_1 and \lim_{x\to a} g(x) = L_2, we know that \forall \epsilon >0 \exists \delta >0 \forall x (0 < |x-a|<\delta \to |f(x)-L_1|<\epsilon) and \forall \epsilon >0 \exists \delta >0 \forall x (0 < |x-a|<\delta \to |g(x)-L_2|<\epsilon) Now let \epsilon > 0. We need to find \delta > 0 such that 0<|x-a|<\delta \to |(f(x)+g(x))-(L_1+L_2)|<\epsilon But \begin{align} |f(x)+g(x)-(L_1+L_2)| &= |f(x)-L_1 + g(x)-L_2| \\ &\leq |f(x)-L_1| + |g(x)-L_2| \end{align} by the triangle inequality. So if we can show |f(x)-L_1| + |g(x)-L_2|<\epsilon, then our result follows immediately. But, \frac{\epsilon}{2} being a positive real number, we have \delta_1,\delta_2 > 0 such that 0 < |x-a|<\delta_1 \to |f(x)-L_1|<\frac{\epsilon}{2} \\ 0 < |x-a|<\delta_2 \to |f(x)-L_2|<\frac{\epsilon}{2} So if we let \delta := \min(\delta_1,\delta_2), then |f(x)-L_1| and |g(x)-L_2| are both less than \frac{\epsilon}{2}, so |f(x)-L_1| + |g(x)-L_2|< \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon which completes the proof. \square


Tags: analysis, math.

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