Compute $\lim_{x \to +\infty}{\frac{\log(a+be^x)}{\sqrt{a+bx^2}}}$

Question

Please help me do the following problem which is in the book Calculus by Tom M. Apostol (vol.1, 2nd edition, p.303, exercise 4.).

Compute $$\lim_{x \to +\infty}{\frac{\log(a+be^x)}{\sqrt{a+bx^2}}}$$ where $a$ and $b$ are positive constants.

L'Hospital's rule for $\infty/\infty$ is not allowed, but I can use the property:$$\lim_{x \to +\infty}{\frac{(\log x)^b}{x^a}}=0$$where $a$ and $b$ are positive constants.

My Attempt:

\begin{align} \frac{\log(a+be^x)}{\sqrt{a+bx^2}} &<\frac{\log(a+be^x)}{\sqrt{bx^2}}\\ &=b^{-1/2}\frac{\log(a+be^x)}{x}\\ &=b^{-1/2}\frac{\log(a+bt)}{\log t}\;\;\;\;\;\;\;(t=e^x)\\ \end{align} then I can't go any more.

Thank you for helping.

Answer 1

For $b>0$ we obtain: $$\lim_{x\rightarrow+\infty}\frac{\ln(a+be^x)}{\sqrt{a+bx^2}}=\lim_{x\rightarrow+\infty}\frac{x+\ln\left(\frac{a}{e^x}+b\right)}{\sqrt{a+bx^2}}=\lim_{x\rightarrow+\infty}\frac{1+\frac{\ln\left(\frac{a}{e^x}+b\right)}{x}}{\sqrt{\frac{a}{x^2}+b}}=\frac{1}{\sqrt{b}}$$

Answer 2

$\ln(a + be^x) \sim \ln(be^x), x \to \infty$ and $\sqrt{a + bx^2} \sim \sqrt{b}x, x \to \infty$ hence your lim is equal to $\frac{1}{\sqrt{b}}$

Answer 3

$$ \lim_{x\to\infty}\frac{\log(a+be^x)}{\sqrt{a+bx^2}} =\lim_{x\to\infty}\frac{\log(e^x)+\log(ae^{-x}+b)}{\sqrt{a+bx^2}} =\lim_{x\to\infty}\left(\frac{x}{\sqrt{a+bx^2}}+\frac{\log(ae^{-x}+b)}{\sqrt{a+bx^2}}\right) =\lim_{x\to\infty}\left(\frac{x}{\sqrt{a+bx^2}}+\frac{\log(ae^{-x}+b)}{ae^{-x}}\cdot\frac{ae^{-x}}{\sqrt{a+bx^2}}\right) =\lim_{x\to\infty}\left(\sqrt{\frac{x^2}{a+bx^2}}+\frac{\log(ae^{-x}+b)}{ae^{-x}}\cdot\frac{a}{e^x\sqrt{a+bx^2}}\right)=\sqrt{\frac{1}{b}}+b\cdot0=\frac{1}{\sqrt{b}} $$