1. ## Limit

any tips on how to do this:

lim(n->infinity) (2^n-5^n)/(3^n+5^n)

thanks

2. Hello

You can try to factor both the numerator and the denominator by $\displaystyle 5^n$.

3. Thanks! got it

4. Originally Posted by ah-bee
any tips on how to do this:

lim(n->infinity) (2^n-5^n)/(3^n+5^n)

thanks
Always try to compare the "top" and "bottom" of a fraction. If the "bottom" is always more than the "top" and either the "top" is decreasing and/or the "bottom" is increasing, then the limit of such a fraction is zero.

"Top" = $\displaystyle 2^n - 5^n$
"Bottom" = $\displaystyle 3^n + 5^n$

Let's see if:

"Top" < "Bottom"
$\displaystyle 2^n - 5^n < 3^n + 5^n$

$\displaystyle 2^n -3^n< 2*5^n$

Note that $\displaystyle 3^n > 2^n$, so the left side yields a negative result. This inequality is always true.

Therefore, we can safely say that the limit is zero. You could have easily just looked at the fraction and made this safe bet.

5. Originally Posted by colby2152
Always try to compare the "top" and "bottom" of a fraction. If the "bottom" is always more than the "top" and either the "top" is decreasing and/or the "bottom" is increasing, then the limit of such a fraction is zero.

"Top" = $\displaystyle 2^n - 5^n$
"Bottom" = $\displaystyle 3^n + 5^n$

Let's see if:

"Top" < "Bottom"
$\displaystyle 2^n - 5^n < 3^n + 5^n$

$\displaystyle 2^n -3^n< + 10^n$

Note that $\displaystyle 3^n > 2^n$, so the left side yields a negative result. This inequality is always true.

Therefore, we can safely say that the limit is zero. You could have easily just looked at the fraction and made this safe bet.
$\displaystyle 2^n - 5^n < 3^n + 5^n$

$\displaystyle 2^n - 3^n < 2\cdot 5^n$ (also, this is not $\displaystyle 10^n$)

You are not allowed to add $\displaystyle 5^n$ to both sides. Doing such thing means,

$\displaystyle \lim_{x\to \infty} \frac{f(x)}{g(x)} = \lim_{x\to \infty}\frac{f(x)+h(x)}{g(x)+h(x)}$ which is usually wrong.

Also, $\displaystyle 2^n - 5^n < 3^n + 5^n$ doesn't prove anything. LHS is approaching $\displaystyle -\infty$ and RHS is approaching $\displaystyle +\infty$ but who knows what their division will be?

The easy way to do these questions is comparing the terms.

$\displaystyle 1<2<...<n<x<2x<3x<...<x^2<x^2<x^n<...<x!<x^x$

The lesser terms will vanish.

So,
$\displaystyle \lim_{n\to \infty} \frac{2^n - 5^n}{3^n+5^n}$

Those $\displaystyle 5^n$ terms will grow much faster than 2^x and 3^x, making them meaningless. So we cross them out.

$\displaystyle \lim_{n\to \infty} \frac{\not 2^n - 5^n}{\not 3^n+5^n}$

$\displaystyle \lim_{n\to \infty} \frac{- 5^n}{5^n}$

$\displaystyle \boxed{-1}$

This was the term comparison method.

The classic approach to these limits (as flyingsquirrel said) is dividing all terms by the biggest term.

$\displaystyle \lim_{n\to \infty} \frac{2^n - 5^n}{3^n+5^n}$

$\displaystyle \lim_{n\to \infty} \frac{\frac{2^n}{5^n} - \frac{5^n}{5^n}}{\frac{3^n}{5^n}+\frac{5^n}{5^n}}$

We know that $\displaystyle \lim_{x\to \infty}r^x = 0$ for $\displaystyle |r|<1$. So $\displaystyle \lim_{n\to \infty} \frac{2^n}{5^n} = \lim_{n\to \infty} \frac{3^n}{5^n} = 0$

$\displaystyle \lim_{n\to \infty} \frac{0 - 1}{0+1} = \boxed{-1}$

6. Originally Posted by wingless
$\displaystyle 2^n - 5^n < 3^n + 5^n$

$\displaystyle 2^n - 3^n < 2\cdot 5^n$ (also, this is not $\displaystyle 10^n$)
Thanks, I have no clue where that 10 came from... I fixed it!