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Find the limit:

$$\lim_{n\to\infty} \frac{(10+3\cos n)^n+6^n}{(10+3\cos n)^n+3^n}$$

My attempt: $1\leq \frac{(10+3\cos n)^n+6^n}{(10+3\cos n)^n+3^n}\leq\frac{(10+3\cos n)^n+6^n}{(10+3\cos n)^n}=1+\frac{6^n}{(10+3\cos n)^n}$

Now I understand that $\frac{6^n}{(10+3\cos n)^n}$ goes to $0$ so by the sandwich theorem I would get that the solution is $1$. However I saw in the solution manual that they wrote $1+\frac{6^n}{(10+3\cos n)^n}\leq 1+\frac{6^n}{7^n}$ and then they said $\frac{6^n}{7^n}$ goes to $0$. Can someone explain how they got $7^n$ in the denominator?

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    $10+3\cos n\ge 10-3\cdot 1=7$. – 2017-01-05

1 Answers 1

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Note that $-1 \le \cos n \le 1$, hence $$ 7 \le 10 + 3\cos n \le 13 $$ Now we get $$ 7^n \le (10 + 3\cos n)^n \le 13^n $$ that gives $$ \frac 1{7^n} \ge \frac 1{(10 + 3\cos n)^n} $$ and therefore $$ \frac {6^n}{(10 + 3\cos n)^n} \le \frac{6^n}{7^n}. $$