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I was trying to solve this problem about limit and I have some problems. $$\lim_{n\rightarrow \infty}(3^n+7^n)^{1/n}$$ I need some help with this limit please.

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    $\begingroup$ To add some colloquialisms to @user17762's answer: for really large $n$, $3^n$ is going to be incredibly small compared to $7^n$. So for large $n$, you can somewhat neglect the $3^n$ part, which is why you get $7$. The appropriate mathematical justification for this is exactly as user17762 has explained. I just wanted to provide some way for you to intuitively arrive at the answer (which can help guide you to figuring out the right argument). $\endgroup$ – Cameron Williams Apr 20 '15 at 20:03
  • $\begingroup$ math.stackexchange.com/questions/1079634/… $\endgroup$ – Bumblebee Apr 24 '15 at 6:51
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HINT: We have $$7^n < 3^n + 7^n < 2\cdot 7^n$$

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$$ (3^n+7^n)^{\frac1n}=\left[7^n\left(\frac{3^n}{7^n}+1\right)\right]^{\frac{1}{n}}=7(1+q^n)^\frac{1}{n} \quad \forall n\in \mathbb{N} $$ with $q=\frac37<1$. Taking the limit we get $$ \lim_n(3^n+7^n)^{\frac1n}=\lim_n7(1+q^n)^\frac{1}{n}=7. $$

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The sandwich lemma states that for any sequence $(a_n)$ with $(b_n)$, $(c_n)$ such that $b_n < a_n < c_n$ for any given natural number $n$, then for any $x$:

$$b_n \to x \text{ and } c_n \to x \implies a_n \to x$$

If for any given $n$, $a_n$ is $(3^n + 7^n)^{1/n}$, you need to find suitable sequences for $(b_n)$ and $(c_n)$ with the same limit as $n \to \infty$ which "sandwich" $(a_n)$.

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$$\lim_{n\rightarrow \infty}(3^n+7^n)^{1/n}=7\lim_{n\rightarrow \infty}\left(\left(\frac37\right)^n+1\right)^{1/n}.$$

The limit on the right is $1$ as

$$1<\left(\left(\frac37\right)^n+1\right)^{1/n}<\left(\frac37\right)^n+1.$$

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