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$$\lim_{x\rightarrow\infty}\frac{5^x}{6^x}$$
 A: 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.
A: Here is a formal answer:
Let $\epsilon>0$. Note that since $0 \leq \frac{(5/6)^n}{(5/6)^{n-1}}=5/6<1$, the sequence is decreasing. We show that there exists some $N \in \mathbb{N}$ so that $n \geq N \implies (\frac{5}{6})^n<\epsilon$. Let $N>\frac{\log(\epsilon)}{\log(5/6)}$, then $\frac{5}{6}^N<(\frac{5}{6})^{\log(\epsilon)/\log(5/6)}=\epsilon.$

Here is an un-aesthetic answer:
 Note that the function $\frac{d}{dx}\frac{5}{6}^x=\ln(5/6)\cdot(\frac{5}{6})^n<0$ so the sequence is decreasing. Also observe that $(\frac{5}{6})^n \geq 0$, so we know the sequence converges.
Suppose that $\lim_{n \to \infty} (\frac{5}{6})^n=L>0$. But we know that since $n=\log(L)/\log(5/6)$ implies equality, we have a contradiction.

Here is a quicker one:
Note that since $\log$ is continuous, we can distribute through the limit in the following way:
$$\log\left(\lim_{n \to \infty} \left(\frac{5}{6}\right)^n\right)=\lim_{n \to \infty} \log\left(\lim_{n \to \infty} \left(\frac{5}{6}\right)^n\right)=\lim_{n \to \infty}n\cdot \log(5/6)\to -\infty$$
$$ \implies 0=\exp \log\left(\lim_{n \to \infty} \left(\frac{5}{6}\right)^n\right)=\lim_{n \to \infty} (\frac{5}{6})^n.$$
A: Let's remember one of the things from high school:
$$ \frac{a^n}{b^n} = \left( \frac{a}{b} \right) ^n $$
Now, what happens when you keep multiplying a number less than 1 by itself over and over?
A: $f(n)=(\frac{5}{6})^n$ is always positive and decreasing. So it must decrease to $0$ or some positive bound $M$, a finite number as $f(0)=1$. By l'hospitals rule and the properties of limits we have:
$$L=\lim_{n \to \infty} (\frac{5}{6})^n=\frac{\ln 5}{\ln 6}L$$
As the limit exists (is finite) and we have the equation $L=\frac{\ln 5}{\ln 6}L$, we must have $L=0$.
