<s>[INST] Limit of $\frac{5^x}{6^x}$ at infinity How should I calculate the limit of the given function? 
Should I differentiate or should I account for changes intuitively? 
Please provide a general answer accounting for both:


*

*when numerator is smaller and 

*when numerator is greater$$\lim_{x\rightarrow\infty}\frac{5^x}{6^x}$$
 [/INST] Note that $\bigl(\frac{5}{6} \bigr)^x$ is a decreasing function of $x$, because  $\bigl(\frac{5}{6} \bigr)^x = e^{x \log (\frac{5}{6})}$, and as $x$ increases, $x \log(\frac{5}{6})$  decreases because $\log(\frac{5}{6}) < 0$, and hence $e^{x \log(\frac{5}{6})}$ is a decreasing function of $x$ for $x>0$ because the exponent then tends to $-\infty$. Now, because $\frac{5}{6}$ is a positive quantity, it follows that $\bigl(\frac{5}{6}\bigr)^x$ is bounded below by zero.
Hence, $\bigl(\frac{5}{6}\bigr)^x$ is a decreasing sequence which is bounded below. By completeness, $\bigl(\frac{5}{6}\bigr)^x$  is a convergent sequence, say to some $C$. Now, because $C = \displaystyle\lim_{x \to \infty} \biggl(\frac{5}{6}\biggr)^x$, it follows that:
$$
\frac{5}{6} C = \displaystyle\lim_{x \to \infty} \biggl(\frac{5}{6}\biggr)^{x+1} = \displaystyle\lim_{x \to \infty} \biggl(\frac{5}{6}\biggr)^x = C
$$  
because if $x+1 \to \infty$, then $x \to \infty$ and vice versa. Hence $\frac{C}{6} = 0$ and $C=0$. Which is to say, $\displaystyle\lim_{x \to \infty} \biggl(\frac{5}{6}\biggr)^x=0$.
There was nothing special about $\frac{5}{6}$ here, any proper fraction would do.
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