please use the Sandwich Theorem to solve the question 1

Q: please use the Sandwich Theorem to solve the question 1

To solve the limit \[ \lim_{n \rightarrow \infty} \sqrt[n]{3^{n} + 4^{n} + 5^{n}} \] using the Sandwich Theorem (also known as the Squeeze Theorem), we start by analyzing the expression inside the root. 1. **Identify the dominant term**: As $ n $ approaches infinity, the term $ 5^{n} $ will dominate the other terms $ 3^{n} $ and $ 4^{n} $. Therefore, we can establish the following inequalities: \[ 5^{n} \leq 3^{n} + 4^{n} + 5^{n} \leq 3^{n} + 4^{n} + 5^{n} \] 2. **Taking the $ n $-th root**: Now, we take the $ n $-th root of all parts of the inequality: \[ \sqrt[n]{5^{n}} \leq \sqrt[n]{3^{n} + 4^{n} + 5^{n}} \leq \sqrt[n]{3^{n} + 4^{n} + 5^{n}} \] 3. **Simplifying the bounds**: The left side simplifies to: \[ \sqrt[n]{5^{n}} = 5 \] For the right side, we can factor out $ 5^{n} $ from the expression inside the root: \[ \sqrt[n]{3^{n} + 4^{n} + 5^{n}} = \sqrt[n]{5^{n} \left( \left( \frac{3}{5} \right)^{n} + \left( \frac{4}{5} \right)^{n} + 1 \right)} = 5 \sqrt[n]{\left( \frac{3}{5} \right)^{n} + \left( \frac{4}{5} \right)^{n} + 1} \] 4. **Evaluating the limit**: As $ n $ approaches infinity, both $ \left( \frac{3}{5} \right)^{n} $ and $ \left( \frac{4}{5} \right)^{n} $ approach 0. Therefore, we have: \[ \sqrt[n]{\left( \frac{3}{5} \right)^{n} + \left( \frac{4}{5} \right)^{n} + 1} \rightarrow 1 \] Thus, we can conclude: \[ 5 \leq \lim_{n \rightarrow \infty} \sqrt[n]{3^{n} + 4^{n} + 5^{n}} \leq 5 \] By the Sandwich Theorem, we find: \[ \lim_{n \rightarrow \infty} \sqrt[n]{3^{n} + 4^{n} + 5^{n}} = 5 \] So, the final answer is: \[ \lim_{n \rightarrow \infty} \sqrt[n]{3^{n} + 4^{n} + 5^{n}} = 5 \]

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To solve the limit $$ \lim_{n \rightarrow \infty} \sqrt[n]{3^{n} + 4^{n} + 5^{n}} $$ using the Sandwich Theorem (also known as the Squeeze Theorem), we start by analyzing the expression inside the root. 1. **Identify the dominant term**: As $ n $ approaches infinity, the term $ 5^{n} $ will dominate the other terms $ 3^{n} $ and $ 4^{n} $. Therefore, we can establish the following inequalities: $$ 5^{n} \leq 3^{n} + 4^{n} + 5^{n} \leq 3^{n} + 4^{n} + 5^{n} $$ 2. **Taking the $ n $-th root**: Now, we take the $ n $-th root of all parts of the inequality: $$ \sqrt[n]{5^{n}} \leq \sqrt[n]{3^{n} + 4^{n} + 5^{n}} \leq \sqrt[n]{3^{n} + 4^{n} + 5^{n}} $$ 3. **Simplifying the bounds**: The left side simplifies to: $$ \sqrt[n]{5^{n}} = 5 $$ For the right side, we can factor out $ 5^{n} $ from the expression inside the root: $$ \sqrt[n]{3^{n} + 4^{n} + 5^{n}} = \sqrt[n]{5^{n} \left( \left( \frac{3}{5} \right)^{n} + \left( \frac{4}{5} \right)^{n} + 1 \right)} = 5 \sqrt[n]{\left( \frac{3}{5} \right)^{n} + \left( \frac{4}{5} \right)^{n} + 1} $$ 4. **Evaluating the limit**: As $ n $ approaches infinity, both $ \left( \frac{3}{5} \right)^{n} $ and $ \left( \frac{4}{5} \right)^{n} $ approach 0. Therefore, we have: $$ \sqrt[n]{\left( \frac{3}{5} \right)^{n} + \left( \frac{4}{5} \right)^{n} + 1} \rightarrow 1 $$ Thus, we can conclude: $$ 5 \leq \lim_{n \rightarrow \infty} \sqrt[n]{3^{n} + 4^{n} + 5^{n}} \leq 5 $$ By the Sandwich Theorem, we find: $$ \lim_{n \rightarrow \infty} \sqrt[n]{3^{n} + 4^{n} + 5^{n}} = 5 $$ So, the final answer is: $$ \lim_{n \rightarrow \infty} \sqrt[n]{3^{n} + 4^{n} + 5^{n}} = 5 $$

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