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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?

3 Answers 3

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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, $2^{1/n}\to 1$.

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    Bleh, I should have seen this; I always try sledgehammers first :) (+1 btw.)2012-04-04
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    @David Mitra: A sledgehammer is a very good tool.2012-04-04
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    whats a sledgehammer? Could you also explain the answer a bit more I don't fully understand how you got there2012-04-04
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    How did you use the squeeze theorem in your example?2012-04-04
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    also would i use the squeeze theorem again for the second question i mentioned above?2012-04-04
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    What second question? To finish the proof by Squeeze Theorem, the left stays at $6$, the part $2^{1/n}(6)$ approaches $6$, and so the term in between, which is what we are interested in, approaches $6$ (look at reference given).2012-04-04
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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 $$ \lim_{n\rightarrow\infty} {100(n+1)+25+6^{n }\over100n+25+6^{n } } $$ 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.

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    Using $\lim \frac{a_{n+1}}{{a_n}}$ is definitely not a sledgehammer.2012-04-04
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    I don't know what L'hopitals rule is so I don't think I can use that. but the first method seems pretty solid. I'm gonna check that out2012-04-04
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    @Jeremy I think you should use Andre's approach.2012-04-04
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Just factor out the dominant term:

$$\lim_{n \to \infty} \sqrt[n] {100n+25+6^n}= \lim_{n \to \infty} (6\sqrt[n] {\frac{100n}{6^n}+\frac{25}{6^n}+1})=6$$

(noting that $\frac{n}{6^n}$ and $\frac{1}{6^n}$ tend to $0$ as ${n \to \infty}$ and that the continuous $n$th-root function preserves limits).

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    If my answer is wrong, will someone please correct/explain it? Thanks! (I don't understand the negative voting.)2012-08-26
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    It is not wrong at all. It is awkward that you say "are basic null sequences **so** they tend to $0$ as $n\to\infty$." I take it you define a null sequence as one with $\lim a_n=0$; so you're saying "They tend to zero so they tend to zero." Maybe it would be better to say "They are null sequences, **meaning** they tend to $0$".2013-01-10
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    @PeterTamaroff Thanks, it reads awkward to me too, so I've duely edited my answer.2013-01-11