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\rightarrow\infty}\frac{17(2^{n}) + 11}{13(4^{n})}=\lim_{n\rightarrow\infty}\frac{17(2^{n})+11}{13(2^{2n})}). Let (t = 2^{n}), as (n\rightarrow\infty), (t\rightarrow\infty). The limit becomes (\lim_{t\rightarrow\infty}\frac{17t + 11}{13t^{2}}).

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

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

Step3: Use limit properties

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

Answer:

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