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Compute the limit of the given function

Evaluate the limit.

    \[ \lim_{x \to +\infty} \frac{\log (a + be^x)}{\sqrt{a+bx^2}}. \]


First, we pull an e^x out of (a+be^x) and an x^2 out of a+bx^2 to write

    \[ \frac{\log(a+be^x)}{\sqrt{a+bx^2}} = \frac{\log((e^x)(ae^{-x} + b)}{x \sqrt{ax^{-2} + b}} = \frac{x + \log(ae^{-x} + b)}{x \sqrt{ax^{-2} + b}}. \]

Then we have

    \begin{align*}  \lim_{x \to +\infty} \frac{\log (a + be^x)}{\sqrt{a+bx^2}} &= \lim_{x \to +\infty} \frac{x + \log(ae^{-x} +b)}{x\sqrt{ax^{-2} + b}} \\[9pt]  &= \lim_{x \to +\infty} \left( \frac{1}{\sqrt{ax^{-2} + b}} + \frac{\log(ae^{-x} + b)}{x \sqrt{ax^{-2} + b}} \right)\\[9pt]  &= \frac{1}{\sqrt{b}} + 0 &(\text{Theorem 7.11})\\  &= \frac{1}{\sqrt{b}}. \end{align*}

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