Answer:
[tex]\displaystyle \lim_{n\rightarrow \infty}\left[\frac{1+(-3)^n}{2^n-(-3)^n}\right]=-1[/tex]
Step-by-step explanation:
Recall that when we are evaluating limits to infinity and negative infinity only, terms with a degree less than the highest degree are insignificant. Therefore, the 1 in the numerator is irrelevant as [tex]n[/tex] approaches infinity:
[tex]\displaystyle \lim_{n\rightarrow \infty}\left[\frac{1+(-3)^n}{2^n-(-3)^n}\right]=\lim_{n\rightarrow \infty}\left[\frac{(-3)^n}{2^n-(-3)^n}\right][/tex]
Multiple by [tex]\displaystyle \frac{\frac{1}{(-3)^n}}{\frac{1}{(-3)^n}}[/tex]:
[tex]\displaystyle \lim_{n\rightarrow \infty}\left[\frac{(-3)^n}{2^n-(-3)^n}\cdot \frac{\frac{1}{(-3)^n}}{\frac{1}{(-3)^n}}\right][/tex]
The numerator is [tex]\displaystyle \frac{(-3)^n}{(-3)^n}=1[/tex] and the denominator is [tex]\displaystyle \frac{2^n-(-3)^n}{(-3)^n}[/tex]. Simplify using [tex]\displaystyle \frac{a-b}{c}=\frac{a}{c}-\frac{b}{c}[/tex]:
[tex]\displaystyle \frac{2^n-(-3)^n}{(-3)^n}=\frac{2^n}{(-3)^n}-\frac{(-3)^n}{(-3)^n}=\frac{2^n}{(-3)^n}-1[/tex]
Thus we have:
[tex]\displaystyle \lim_{n\rightarrow \infty}\left[\frac{(-3)^n}{2^n-(-3)^n}\cdot \frac{\frac{1}{(-3)^n}}{\frac{1}{(-3)^n}}\right]=\lim_{n\rightarrow \infty}\left[ \frac{1}{\frac{2^n}{(-3)^n}-1}\right][/tex]
The limit of [tex]\displaystyle \lim_{n\rightarrow \infty} \frac{2^n}{(-3)^n}[/tex] is simply 0 because the denominator grows faster than the numerator. The value varies from negative to positive as [tex]n[/tex] varies from being odd or even respectively, but it still approaches zero from both sides. Graphing might be unclear because the exponent in the denominator has a negative base, so [tex]n[/tex] is only defined for all integers ([tex]n\in \mathbb{Z}[/tex]).
Hence, we have:
[tex]\displaystyle \lim_{n\rightarrow \infty}\left[\frac{(-3)^n}{2^n-(-3)^n}\cdot \frac{\frac{1}{(-3)^n}}{\frac{1}{(-3)^n}}\right]=\lim_{n\rightarrow \infty}\left[ \frac{1}{\frac{2^n}{(-3)^n}-1}\right]=\frac{1}{0-1}=\frac{1}{-1}=\boxed{-1}[/tex]
