How to evaluate $\lim \limits_{n\rightarrow \infty}(n+9)^{\frac{1}{n}}$? I have no idea where to begin to evaluate  $\lim \limits_{n\rightarrow \infty}(n+9)^{\frac{1}{n}}$.
The given answer is $1$. I do know that $\lim \limits_{n\rightarrow \infty} n^{\frac{1}{n}}=1$ (my text states this is an important result to remember). I've tried to treat this as the composition 2 two functions and evaluate them accordingly but that yield $\infty^0$, which is undefined. So there must be a trick I need to do to rewrite this expression, but I don't know that it is.
 A: Take $\log$:
$$
\log L = \lim_{n\to\infty}\log(n+9)^{\frac{1}{n}} =
\lim_{n\to\infty}\frac{\log(n+9)}n = \cdots
$$
A: There is a general theorem that says if $\lim f(u)$ and $\lim g(u)$ both exist, then so does $\lim(f(u)g(u))$ and
$$\lim(f(u)g(u))=\left(\lim f(u)\right)\!\!\left(\lim g(u)\right)$$
where the variable $u$ can be either a discrete $n$ tending to infinity or a real $x$ tending to a point on the real line (or to infinity).  For the problem at hand,
$$(n+9)^{1/n}=n^{1/n}\left(1+{9\over n}\right)^{1/n}$$
Since $\lim_{n\to\infty}n^{1/n}=1$ is stipulated as known and $\lim_{n\to\infty}(1+9/n)^{1/n}=1$ is obvious (or at least easy to see), the result follows.
Remark:  When we say a limit "exists" here, we mean it exists as a real number.  Otherwise we could wind up in a $\infty\!\cdot\!0$ situation.
A: $$ n^{\frac{1}{n}} \leq (n+9)^{\frac{1}{n}} \leq (2n)^\frac{1}{n}$$
For $ n $ greater than, say, $ 9 $. Apply "squeeze theorem"
A: My guideline is to avoid power notation for non-integer exponents. 
Rewrite
$$
(9+n)^{1/n} = \exp \frac 1n\log (9+n)
$$
You should know that the limit of $\frac {\log (a+bn)}n$ is zero. Conclude using the continuity.
A: Squeezing is best. However, we can also use the fact that
$$(n+9)^{\frac{1}{n}}=\left( (n+9)^{\frac{1}{n+9}}  \right)^{\frac{n+9}{n}}$$
A: The basic inequality needed
is Bernoulli's inequality:
If $x \ge 0$ and
$n$ is a positive integer
then
$(1+x)^n
\ge 1+nx
$.
This is easily proved by induction.
From this,
putting
$\frac{x}{n}$ for $x$,
$(1+\frac{x}{n})^n
\ge 1+x
$.
Taking the $n^{th}$ root,
$(1+x)^{1/n}
\le 1+\frac{x}{n}
$.
Now,
we can consider your problem.
$(n+9)^{1/n}
=n^{1/n}(1+\frac{9}{n})^{1/n}
\le n^{1/n}(1+\frac{9}{n^2})
$.
To show that
$n^{1/n} \to 1$,
put $x = n^{-1/2}$
in
$(1+x)^n
\ge 1+nx
$.
We get
$(1+n^{-1/2})^n
\ge 1+nn^{-1/2}
=1+n^{1/2}
>n^{1/2}
$.
Raising both sides
to the
$2/n$ power,
this becomes
$(1+n^{-1/2})^2> n^{1/n}$
or
$n^{1/n}
<(1+n^{-1/2})^2
=1+2n^{-1/2}+1/n
\le 1+\frac{3}{\sqrt{n}}
$.
Therefore,
$(n+9)^{1/n}
\le n^{1/n}(1+\frac{9}{n^2})
< (1+\frac{3}{\sqrt{n}})(1+\frac{9}{n^2})
$.
Presto,
no logs and all constants explicit.
