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Find a constant so that a given limit formula holds

Find a value for the constant c so that the following formula is true,

    \[ \lim_{x \to +\infty} \left( \frac{x+c}{x-c} \right)^x = 4. \]


Using the definition and continuity of the exponential and then L’Hopital’s rule we solve for the constant,

    \begin{align*}  &&\lim_{x \to +\infty} \left( \frac{x+c}{x-c} \right)^x &= 4 \\[9pt]  \implies && \lim_{x \to +\infty} e^{ x (\log (x+c) - \log(x-c))} &= 4\\[9pt]  \implies && \exp \left( \lim_{x \to +\infty} x (\log(x+c) - \log(x-c)) \right) &= 4 \\[9pt]  \implies && \lim_{x \to +\infty} x(\log(x+c) - \log(x-c)) &= \log 4 \\[9pt]  \implies && \lim_{x \to +\infty} \frac{\log(x+c) - \log(x-c)}{\frac{1}{x}} &= 2 \log 2 \\[9pt]  \implies && \lim_{x \to +\infty} \frac{ \frac{1}{x+c} - \frac{1}{x-c}}{-\frac{1}{x^2}} &= 2 \log 2 \\[9pt]  \implies && \lim_{x \to +\infty} \left( \frac{x^2}{x-c} - \frac{x^2}{x+c} \right) &= 2 \log 2 \\[9pt]  \implies && \lim_{x \to +\infty} \frac{x^3 + cx^2 - x^3 + cx^2}{(x-c)(x+c)} &= 2 \log 2 \\[9pt]  \implies && \lim_{x \to +\infty} \frac{ 2cx^2}{x^2 - c^2} &= 2 \log 2 \\[9pt]  \implies && \lim_{x \to +\infty} \frac{2c}{1 - \left( \frac{c}{x} \right)^2} &= 2 \log 2 \\[9pt]  \implies && 2c &= 2\log 2 \\[9pt]  \implies && c &= \log 2.  \end{align*}

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