1. ## Question on Limits

Evaluate $\lim_{x \to \infty} \frac{(\frac{10}{11})^x}{(\frac{9}{10})^x + (\frac{11}{12})^x}$

Instinctively it's $\frac{1}{2}$, just that I'm not too sure how to go about showing it.

2. ## Re: Limit

Note that 9/10, 10/11, and 11/12 are all less than one, so $\left(\frac{9}{10}\right)^x$, $\left(\frac{10}{11}\right)^x$, and $\left(\frac{11}{12}\right)^x$ all tend to zero as x goes to infinity. The largest of these numbers is 11/12, so it goes to zero most slowly. Thus, we multiply both the numerator and denominator by $\left(\frac{12}{11}\right)^x$:
$\lim_{x\to\infty}\frac{(\frac{10}{11})^x}{(\frac{9 }{10})^x+(\frac{11}{12})^x}=\lim_{x\to\infty}\frac {(\frac{10}{11})^x(\frac{12}{11})^x}{((\frac{9}{10 })^x+(\frac{11}{12})^x)(\frac{12}{11})^x}$
$=\lim_{x\to\infty}\frac{(\frac{120}{121})^x}{(\fra c{54}{55})^x+1}$.
Note that 120/121 and 54/55 are both less than one, as expected, and so as x goes to infinity, the numerator of the above goes to zero, and the denominator goes to one. Thus the limit is 0.

--Kevin C.

3. $\lim_{x\to\infty} \frac{\left ( \frac{10}{11} \right )^x}{\left ( \frac{9}{10} \right )^x + \left ( \frac{11}{12} \right )^x}$

The general approach for these limits is factoring the largest term of the denominator. It is $\left ( \frac{11}{12} \right )^x$ here.

$\lim_{x\to\infty} \frac{\left ( \frac{11}{12} \right )^x \cdot \left ( \frac{120}{121} \right )^x}{\left ( \frac{11}{12} \right )^x \cdot \left [ 1 + \left ( \frac{108}{110} \right )^x \right ] }$

$\lim_{x\to\infty} \frac{\left ( \frac{120}{121} \right )^x}{1 + \left ( \frac{108}{110} \right )^x }$

We know that $\lim_{x\to\infty} r^x = 0$ for $|r|<1$. So those fractions will go to zero.

$\lim_{x\to\infty} \frac{0}{1+0}=0$

4. Hello
Originally Posted by h2osprey
Evaluate $\lim_{x \to \infty} \frac{(\frac{10}{11})^x}{(\frac{9}{10})^x + (\frac{11}{12})^x}$

Instinctively it's $\frac{1}{2}$, just that I'm not too sure how to go about showing it.
Another possible approach : $0<\frac{(\frac{10}{11})^x}{(\frac{9}{10})^x + (\frac{11}{12})^x}\leq\frac{(\frac{10}{11})^x}{(\f rac{11}{12})^x}=\left(\frac{120}{121}\right)^x\und erset{x\to\infty}{\to}0$ and we can conclude using the squeezing theorem.