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Prove that sequences with finite limits are linear with respect to taking limits

Let

    \[ \lim_{n \to \infty} a_n = A \qquad \text{and} \qquad \lim_{n \to \infty} b_n = B. \]

Prove directly from the definition of the limit that

    \[ \lim_{n \to \infty} (a_n + b_n) = A + B \qquad \text{and} \qquad \lim_{n \to \infty} (ca_n) = cA \]

for any constant c.


Proof. Let \varepsilon >0 be given. Since \lim_{n \to \infty} a_n = A and \lim_{n \to \infty} b_n = B we know there exist positive integers N_a and N_b such that

    \[ |a_n - A| < \frac{\varepsilon}{2} \qquad \text{and} \qquad |b_m - B| < \frac{\varepsilon}{2} \]

for all n \geq N_a and all m \geq N_b. Let N = \max \{ N_a, N_b \}. Then, for all n \geq N we have

    \begin{align*}  &&|a_n - A| + |b_n - B| &< \varepsilon \\  \implies && |a_n + b_n - (A+B)| &< \varepsilon. \end{align*}

Hence,

    \[ \lim_{n \to \infty} (a_n + b_n) = A+B. \qquad \blacksquare \]

Proof. Let \varepsilon > 0 be given. Since \lim_{n \to \infty} a_n = A we know there exists a positive integer N such that

    \[ |a_n - A| < \frac{\varepsilon}{|c|} \qquad \text{for all } n \geq N. \]

Thus,

    \begin{align*}  &&|c| \cdot |a_n - A| &< |c| \cdot \frac{\varepsilon}{|c|} \\[9pt]  \implies && | ca_n - cA| &< \varepsilon \\[9pt]  \implies && \lim_{n \to \infyt} ca_n &= cA. \qquad \blacksquare \end{align*}

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