Question

lim_{n\to\infty}\frac{17(2^{n}) + 11}{13(4^{n})}

Explanation:

Step1: Rewrite the expression

We know that \(4^{n}=(2^{2})^{n}=2^{2n}\). So the given limit \(\lim_{n
ightarrow\infty}\frac{17(2^{n}) + 11}{13(4^{n})}=\lim_{n
ightarrow\infty}\frac{17(2^{n})+11}{13(2^{2n})}\). Let \(t = 2^{n}\), as \(n
ightarrow\infty\), \(t
ightarrow\infty\). The limit becomes \(\lim_{t
ightarrow\infty}\frac{17t + 11}{13t^{2}}\).

Step2: Divide numerator and denominator by \(t^{2}\)

\(\lim_{t
ightarrow\infty}\frac{\frac{17t}{t^{2}}+\frac{11}{t^{2}}}{\frac{13t^{2}}{t^{2}}}=\lim_{t
ightarrow\infty}\frac{\frac{17}{t}+\frac{11}{t^{2}}}{13}\).

Step3: Use limit properties

We know that \(\lim_{t
ightarrow\infty}\frac{1}{t}=0\) and \(\lim_{t
ightarrow\infty}\frac{1}{t^{2}} = 0\). So \(\lim_{t
ightarrow\infty}\frac{\frac{17}{t}+\frac{11}{t^{2}}}{13}=\frac{\lim_{t
ightarrow\infty}\frac{17}{t}+\lim_{t
ightarrow\infty}\frac{11}{t^{2}}}{13}\). Substituting the limit - values, we get \(\frac{0 + 0}{13}=0\).

Answer:

0