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.
On
$$ n^{\frac{1}{n}} \leq (n+9)^{\frac{1}{n}} \leq (2n)^\frac{1}{n}$$
For $ n $ greater than, say, $ 9 $. Apply "squeeze theorem"
On
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.
On
Take $\log$: $$ \log L = \lim_{n\to\infty}\log(n+9)^{\frac{1}{n}} = \lim_{n\to\infty}\frac{\log(n+9)}n = \cdots $$
On
Squeezing is best. However, we can also use the fact that $$(n+9)^{\frac{1}{n}}=\left( (n+9)^{\frac{1}{n+9}} \right)^{\frac{n+9}{n}}$$
On
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.
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.