Evaluate the limit as n approaches infinity of (2^n + (-7)^n) divided by (2^(n+1) + (-7)^(n+1)).

Step-by-step solution

1. 1

   Find the dominant term by comparing absolute values. The competing bases have sizes $|2|=2$ and $|-7|=7$. Since $7>2$, the terms $(-7)^n$ swamp the terms $2^n$ as $n$ grows — the alternating sign does not change which one is bigger.

2. 2

   Split off one factor from the shifted powers. Rewrite the denominator so both terms match the numerator's exponent:

   $2^{n+1}=2\cdot 2^{n},\qquad (-7)^{n+1}=(-7)\cdot(-7)^{n}.$

3. 3

   Divide numerator and denominator by the dominant term. Let $r=\dfrac{2}{-7}$, so $\dfrac{2^{n}}{(-7)^{n}}=r^{n}$:

   $\frac{2^{n}+(-7)^{n}}{2\cdot 2^{n}-7\cdot(-7)^{n}}=\frac{r^{n}+1}{2r^{n}-7}.$

4. 4

   Send the small ratio to zero. Since $|r|=\tfrac27<1$, $r^{n}\to 0$ as $n\to\infty$ — it alternates in sign but shrinks geometrically, so its limit is $0$ regardless.

5. 5

   Read off the limit.

   $\lim_{n\to\infty}\frac{r^{n}+1}{2r^{n}-7}=\frac{0+1}{0-7}=-\frac{1}{7}.$

6. 6

   Check with exact arithmetic. Evaluating the original quotient with exact fractions gives $-0.14285781$ at $n=10$, $-0.14285695$ at $n=11$, and $-0.142857142857$ at $n=50$, $51$, $200$ and $201$ — converging to $-\tfrac17=-0.142857\ldots$ from both sides as the parity of $n$ alternates.