Limits of convergent/divergent sequences (revision): $\lim\limits_{n \to \infty} \sqrt[n] {100n+25+6^n}$

Question

I need to find the limit :

$$\displaystyle \lim_{n \to \infty} \sqrt[n] {100n+25+6^n}$$

I also got limit ${a^{1/n}} = 1$, ${n^{1/n}}= 1$ and $(1+1/n){^n} = e$

I haven't come across a question like this before so I'm stuck on how to tackle it. My first thoughts are to use the bernoulli inequality since the question I got afterwards is $\lim_{n \to \infty}$ $(1 + \frac{3}{n^2})^{n^2}$ and I obviously can't expand it fully or cancel it out easily.

Any tips?

Answer 1

You could use the result often called the Squeeze Theorem. For large enough $n$ (and it doesn't have to be very large!), we have $100n+25+6^n<2\cdot 6^n$ and therefore $$6 <\sqrt[n]{100n+25+6^n}<2^{1/n}(6).$$ Now let $n \to\infty$. As you mentioned, $2^{1/n}\to 1$.

Answer 2

One way:

Use the fact that for positive $a_n$, if $\lim\limits_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim\limits_{n\rightarrow\infty} \root n\of{a_n}$ and in this case the two limits are equal (see the last page here for a proof of this fact).

For your sequence $$ \lim_{n\rightarrow\infty} {100(n+1)+25+6^{n + 1}\over100n+25+6^{n } } $$ is relatively easy to compute (by dividing each term by $6^n$, for example).

Another way:

First calculate $\lim\limits_{n\rightarrow\infty}\ln\root n\of {100n+25+6^n} =\lim\limits_{n\rightarrow\infty}{\ln ( {100n+25+6^n})\over n}$ using L'Hôpital's rule.