Q:

How does lim $\frac{\ln{ 2^x \over 25^x}}{x} = -\ln \frac{25}{2}$

$$\lim_{x\to\infty}\frac{\ln{ 2^x \over 25^x}}{x} = -\ln \frac{25}{2}$$
How is this answer is obtained?
and How would you work this question ?
$\sum _{n=0}^{\infty }\:\left(\frac{\left(2^n\right)}{25^n}\right)$
thanks

A:

Note that $$\frac{2^x}{25^x} = \left(\frac{2}{25}\right)^x$$
And:
$$\ln \frac{2}{25} = -\ln \frac{25}{2}$$ 
So you don't even need to take the limit. 
For the second, again, $\frac{2^n}{25^n}=\left(\frac{2}{25}\right)^n$ and thus it is a geometric series.

