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Compute the limit of (1+xn)1/n and (an + bn1/n

  1. Consider the limit

        \[ \lim_{n \to \infty} (1+x^n)^{\frac{1}{n}}. \]

    For 0 < x < 1 prove that this limit exists and compute the limit.

  2. For positive real numbers a and b, Consider the limit

        \[ \lim_{n \to \infty} (a^n + b^n)^{\frac{1}{n}}. \]

    Compute this limit.

  1. Proof. We use the squeeze theorem to prove existence and compute the value of the limit. Since 0<x<1 we have

        \[ (1+0^n)^{\frac{1}{n}} < (1+x^n)^{\frac{1}{n}} < (1+1^n)^{\frac{1}{n}} \]

    for all positive integers n. Then we have

        \begin{align*}  \lim_{n \to \infty} (1+0^n)^{\frac{1}{n}} &= \lim_{n \to \infty} 1^{\frac{1}{n}} = 1 \\[9pt]  \lim_{n \to \infty} (1+1^n)^{\frac{1}{n}} &= \lim_{n \to \infty} 2^{\frac{1}{n}} = 1. \end{align*}

    (We know the second limit from this previous exercise (Section 10.4, Exercise #9).) Therefore, by the squeeze theorem we have

        \[ \lim_{n \to \infty} \left( 1 + x^n \right)^{\frac{1}{n}} = 1. \qquad \blacksquare \]

  2. If a > b then we have 0 < \frac{a}{b} < 1 and so, using part (a),

        \begin{align*}  \lim_{n \to \infty} \left( a^n + b^n \right)^{\frac{1}{n}} &= \lim_{n \to \infty} a \cdot \left( 1 + \left( \frac{b}{a} \right)^n \right)^{\frac{1}{n}} \\[9pt]  &= a. \end{align*}

    On the other hand if b > a then we have 0 < \frac{a}{b} < 1 and so by part (a) again,

        \begin{align*}  \lim_{n \to \infty} \left( a^n + b^n \right)^{\frac{1}{n}} &= \lim_{n \to \infty} b \cdot \left( 1 + \left(\frac{a}{b}\right)^n \right)^{\frac{1}{n}} \\[9pt]  &= b. \end{align*}

    If a = b then

        \begin{align*}  \lim_{n \to \infty} (a^n + b^n)^{\frac{1}{n}} &= a \cdot \lim_{n \to \infty} \left( 1 + \left( \frac{b}{a} \right)^n \right)^{\frac{1}{n}} \\[9pt]  &= a \cdot \lim_{n \to \infty} 2^{\frac{1}{n}} \\[9pt]  &= a = b. \end{align*}

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