Limit

Question

Can we calculate.   $\textstyle \lim_{n \to \infty}\frac{2^{n}}{3^{n}}$ if yes then what is the value.    

Answer 1

$\textstyle \lim_{n \to \infty}\frac{2^{n}}{3^{n}}$

=$\textstyle \lim_{n \to \infty}(\frac{2}{3})^{n}$

Since 2/3 is between 0 and 1, the higher the exponent, the lower the result (although the result will always be positive).

So the limit is Zero.

Answer 2

Given $\lim_{n\rightarrow \infty}\frac{2^{n}}{3^{n}}$

$= \lim_{n\rightarrow \infty}\left ( \left ( \frac{2}{3} \right )^{n} \right )$

$= \lim_{n\rightarrow \infty}\left ( e^{n ln\left ( \frac{2}{3} \right )} \right )$

Using $a^{x}=e^{x\ lna}$ the following property,

Apply the chain rule,

$\lim_{u\rightarrow b}f\left ( u \right )= L\ and\ \lim_{x\rightarrow a}g\left ( x \right )\doteq b$

Then $\lim_{x\rightarrow a} f\left ( g\left ( x \right ) \right )= L$

$g\left ( n \right )= nln\left ( \frac{2}{3} \right )$ and $f\left ( u \right )= e^{u}$

So $\lim_{n\rightarrow \infty}\left ( n ln\left ( \frac{2}{3} \right ) \right )= -\infty$

$\lim_{u\rightarrow -\infty }e^{u}= 0$

Comments

given limit is $\textstyle \lim_{n \to \infty}$ not $\textstyle \lim_{n \to 0}$

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Sorry by mistake

Answer 3

$\textstyle \lim_{n \to \infty}\frac{2^{n}}{3^{n}}$

y  = $\textstyle \lim_{n \to \infty}(\frac{2}{3})^{n}$

Taking log both side,
log y = $lim n \to \infty$n log(2/3)

log y =$lim n \to \infty$ n (-.17609)

y =$lim n \to \infty$ e^ (n * (-.17609)) [ Now here in this  step, I have a doubt that if we multiply negative no. with positive infinity then will it change it to negative???] 

If above statement is true then,

y = e ^ (- $\infty$) 

y = 0