# Finding $\lim_{x\to \pm\infty}f(x)$ where $a,b>0$

I found this problem interesting to be here:

If $a,b>0$ then find the following limit: $$\lim_{x\to\pm\infty}\left(\frac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x$$

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#### Solutions Collecting From Web of "Finding $\lim_{x\to \pm\infty}f(x)$ where $a,b>0$"

One can use L’Hospital’s Rule mechanically. The logarithm of our expression is
$$\frac{\log\left(\frac{a^{1/x}+b^{1/x}}{2}\right)}{1/x}.$$
Differentiate top and bottom. We get
$$\frac{1}{\frac{a^{1/x}+b^{1/x}}{2}}(-1/x^2)\frac{\frac{a^{1/x}\log a+b^{1/x}\log b}{2}}{(-1/x^2)}$$
Cancel the $(-1/x^2)$. The limit is $\dfrac{\log a+\log b}{2}$. Then exponentiate.