Limit of function at infinity: $ \lim_{n\to \infty} \frac{a^n-b^n}{a^n+b^n} $ I'm learning the basics of Calculus. I got stuck in a very (seemingly) simple problem of evaluating a limit. The problem given is:
$$
\lim_{n\to \infty} \frac{3^n-2^n}{3^n+2^n}
$$
The answer given is $1$ (I have no idea how, or why).
I think $\lim_{x \to a}\frac{x^m-a^m}{x^n-a^n}=\frac{m}{n}a^{m-n}$, isn't going to help me in any way because in this formula, 


*

*$x \to a$, not $\infty$

*$x$ is not the exponent.


Sadly, there's no other formula in my book that seems close to the problem. So could anyone please show me how to solve such problems? Also, could you show me how to solve the general form of the problem, like:
$$
\lim_{n\to \infty} \frac{a^n-b^n}{a^n+b^n}
$$
Thanks!
 A: The other answers have already shown the standard trick. Here's how we might have proceeded if we hadn't noticed it applies.
One significant point we should understand is that as $n \to \infty$, $3^n$ dwarfs $2^n$ in size; the difference between $3^n$ and $3^n \pm 2^n$ should be negligible. So we estimate
$$ \frac{3^n - 2^n}{3^n + 2^n} \sim 1 $$
Of course, an estimate is not proof: so we look for the error term. We want to write
$$ \frac{3^n - 2^n}{3^n + 2^n} = 1 + x $$
But what is $x$? It's actually easy to find: just solve:
$$ x = \frac{3^n - 2^n}{3^n + 2^n} - 1 = -\frac{2 \cdot 2^n}{3^n + 2^n} $$
and so we want to find
$$ \lim_{n \to \infty} \left( 1 - \frac{2 \cdot 2^n}{3^n + 2^n} \right) $$
We might do this by recognizing that
$$ 0 < \frac{2^n}{3^n + 2^n} < \frac{2^n}{3^n} = \left( \frac{2}{3} \right)^n$$
so that we can use the squeeze theorem. But maybe we didn't notice that, nor did we notice we can use the standard trick. We could still repeat the idea above and estimate $x \sim -2 \cdot (2/3)^n$. Again finding the error term, we want
$$ \lim_{n \to \infty}\left( 1 - 2 \left(\frac{2}{3} \right)^n + \frac{2 \cdot 2^{2n}}{3^n (3^n + 2^n)} \right) $$
or rearranging,
$$ \lim_{n \to \infty}\left( 1 - 2 \left(\frac{2}{3} \right)^n + \left( \frac{2}{3} \right)^n \cdot \frac{2 \cdot 2^n}{3^n + 2^n} \right) $$
Now, the only 'trick' we need to compute the limit is to realize that
$$ 0 < \frac{2 \cdot 2^n}{3^n + 2^n} < 2 $$
because $2^n$ is always smaller than the denominator. Since $(2/3)^n \to 0$, this is enough to know that the third term vanishes.

The standard trick is, actually, not too dissimilar from the idea above. When we estimate $3^n \sim 3^n - 2^n$, the idea is rather than adding in an error term, we multiply in the error term: we want to write
$$ 3^n - 2^n = 3^n \cdot (1 + x) $$
and in this case, we get
$$ 3^n - 2^n = 3^n \cdot \left(1 - \left(\frac{2}{3}\right)^n\right) $$
Another way of thinking about this same idea is that we want to factor out the most significant part: in this case, $3^n$.
A: Hint: divide the numerator and denominator by $3^n$ or if you have the general case , then divide by the max of the two. Namely, if $a > b \to L = \displaystyle \lim_{n\to \infty} \dfrac{1-(\frac{b}{a})^n}{1+(\frac{b}{a})^n} = 1$. The question is the case $a = 3 > 2 = b$.
A: Hint 
For the first part $$\frac{3^n-2^n}{3^n+2^n}=\frac{1-(\frac23)^n}{1+(\frac23)^n}$$
For the second part, consider the two cases : $a >b$ and $a<b$ and divide by the largest.
A: Don't have enough points for a comment, but just to ay that the $3^n$ terms will ,  that the $3^n$ terms will eventually dominate the expression, as $ n \rightarrow \infty$ate 
A: You can rewrite the limit as
$$\lim_{n \to \infty} \frac{3^n - 2^n}{3^n + 2^n} = 
\lim_{n \to \infty} \frac{2\cdot 3^n}{3^n + 2^n} - 1.$$
Since $3^n$ dominates $2^n$ for $n$ large enough, the left hand side will converge to $2$ so you get that the limit is $2-1 = 1$.
