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Prove that the limit of a product is the product of the limits for sequences with finite limits

Let

    \[ \lim_{n \to \infty} a_n = A. \]

From the previous two exercises here and here (Section 10.4, Exercises #30 and #31) we know that

    \[ \lim_{n \to \infty} a_n = 0 \quad \implies \quad \lim_{n \to \infty} a_n^2 = 0 \]

and that

    \[ \lim_{n \to \infty} a_n = A, \ \ \lim_{n \to \infty} b_n = B \quad \implies \quad \lim_{n \to \infty} (a_n + b_n) = A+B. \]

Use these results to prove that

    \[ \lim_{n \to \infty} a_n^2 = A^2. \]

Then use the identity

    \[ 2 a_n b_n = (a_n + b_n)^2 - a_n^2 - b_n^2 \]

to prove that

    \[ \lim_{n \to \infty} (a_n b_n) = AB. \]


Proof. First,

    \begin{align*}  \lim_{n \to \infty} a_n = A && \implies && \lim_{n \to \infty} (a_n - A) &= 0 \\  && \implies && \lim_{n \to \infty} (a_n - A)^2 &= 0 (\text{Ex. } 30)\\  && \implies && \lim_{n \to \infty} (a_n^2 - 2Aa_n + A^2) &= 0 \\  && \implies && \lim_{n \to \infty} a_n^2 - 2A \lim_{n \to \infty} a_n + A^2 &= 0 \\  && \implies && \lim_{n \to \infty} a_n^2 &= A^2. \qquad \blacksquare \end{align*}

Proof. Letting

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

we have from Exercise #31,

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

Then by Exercise #30,

    \[ \lim_{n \to \infty} (a_n + b_n)^2 = (A+B)^2 = A^2 + 2AB + B^2. \]

Using the identity 2a_n b_n = (a_n + b_n)^2 - a_n^2 - b_n^2 we then have

    \begin{align*}  &&\lim_{n \to \infty} (a_n + b_n)^2 &= A^2 + 2AB + B^2 \\  \implies && \lim_{n \to \infty} (2a_n b_n + a_n^2 + b_n^2) &= A^2 + 2AB + B^2 \\  \implies && 2 \lim_{n \to \infty} a_n b_n + \lim_{n \to \infty} a_n^2 + \lim_{n \to \infty} b_n^2 &= A^2 + 2AB + B^2 \\  \implies && 2 \lim_{n \to \infty} a_n b_n &= 2AB \\  \implies && \lim_{n \to \infty} a_n b_n &= AB. \qquad \blacksquare \end{align*}

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