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In two different ways, prove the limit as x tends to 0 of (log (1+x))/x is 1

by RoRi

The following limit equation is valid:

    \[ \lim_{x \to 0} \frac{\log (1+x)}{x} = 1. \]

Prove this in the following two ways:

  1. Using the definition of the derivative L'(1);
  2. Using the previous exercise (Section 6.9, Exercise #28).
  1. Proof. First, we know from Theorem 6.1 part (b) (page 229 of Apostol) that

        \[ L'(x) = \frac{1}{x} \qquad \text{for all } x > 0. \]

    Therefore, L'(1) = 1. Then, we recall the definition of the derivative of a function

        \[ f'(x) = \lim_{x \to 0} \frac{f(x+h) - f(x)}{h} . \]

    So, using the definition of the derivative and the fact that for x > 0 we have L(x) = \log x,

        \begin{align*} L'(x) &= \lim_{h \to 0} \frac{L(x+h) - L(x)}{h} \\ &= \lim_{h \to 0} \frac{\log(x+h) - \log(x)}{h}. \end{align*}

    Evaluating at 1 we have

        \begin{align*} 1 = L'(1) &= \lim_{h \to 0} \frac{\log (1+h) - \log 1}{h} \\ &= \lim_{h \to 0} \frac{\log (1+h)}{h} \\ &= \lim_{x \to 0} \frac{\log (1+x)}{x} \end{align*}

    since the variable name is unimportant. Thus, we have established the requested formula,

        \[ \lim_{x \to 0} \frac{\log (1+x)}{x} = 1. \qquad \blacksquare \]

  2. Proof. From the previous exercise (Section 6.9, Exercise #28) we know

        \[ 1 - \frac{1}{x} < \log x < x - 1 \qquad \text{for all } x > 0 \text{ with } x \neq 1. \]

    If x > 0 then x+1 > 0 as well, so the inequalities still hold with x + 1,

        \[ 1 - \frac{1}{x+1} < \log (x+1) < x. \]

    Multiplying all of the terms by \frac{1}{x} (since x > 0 we may do this without reversing the inequalities),

        \[ \frac{1}{x} - \frac{1}{x(x+1)} < \frac{\log(x+1)}{x} < 1. \]

    Since \frac{1}{x+1} = \frac{1}{x} - \frac{1}{x(x+1)} we then have,

        \[ \frac{1}{x+1} < \frac{\log(x+1)}{x} < 1. \]

    Finally, since

        \[ \lim_{x \to 0} \frac{1}{1+x} = 1 , \qquad \text{and} \qquad \lim_{x \to 0} 1 = 1, \]

    we apply the squeeze theorem (Theorem 3.3, page 133 of Apostol) to conclude,

        \[ \lim_{x \to 0} \frac{\log(1+x)}{x} = 1. \qquad \blacksquare\]

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