Questions: Calculate the limit: lim as n approaches infinity of (sqrt(n+1) - sqrt[3](27n^6+7)) / (sqrt[3](3n^3+1) + sqrt(4n^4+5)) =

Solution Steps

To solve the limit as \( n \to \infty \), we need to analyze the behavior of the numerator and the denominator separately. The dominant terms in the numerator and denominator will determine the limit. Simplify each expression by factoring out the highest power of \( n \) and then evaluate the limit.

The given limit is:

\[ \lim _{n \rightarrow \infty} \frac{\sqrt{n+1}-\sqrt[3]{27 n^{6}+7}}{\sqrt[3]{3 n^{3}+1}+\sqrt{4 n^{4}+5}} \]

Numerator:

\[ \sqrt{n+1} - \sqrt[3]{27n^6 + 7} \]

• The dominant term in \(\sqrt{n+1}\) is \(\sqrt{n}\).
• The dominant term in \(\sqrt[3]{27n^6 + 7}\) is \((27n^6)^{1/3} = 3n^2\).

Denominator:

\[ \sqrt[3]{3n^3 + 1} + \sqrt{4n^4 + 5} \]

• The dominant term in \(\sqrt[3]{3n^3 + 1}\) is \((3n^3)^{1/3} = n\).
• The dominant term in \(\sqrt{4n^4 + 5}\) is \(\sqrt{4n^4} = 2n^2\).

As \( n \to \infty \), the dominant terms will determine the behavior of the expression:

• Numerator: \(\sqrt{n} - 3n^2\)
• Denominator: \(n + 2n^2\)

The limit becomes:

\[ \lim _{n \rightarrow \infty} \frac{\sqrt{n} - 3n^2}{n + 2n^2} \]

Factor out the highest power of \( n \) in both the numerator and the denominator:

\[ = \lim _{n \rightarrow \infty} \frac{n^{1/2}(1 - 3n^{3/2})}{n(1 + 2n)} \]

Simplify:

\[ = \lim _{n \rightarrow \infty} \frac{1 - 3n^{3/2}}{n^{1/2}(1 + 2n)} \]

As \( n \to \infty \), the term \( -3n^{3/2} \) in the numerator and \( 2n \) in the denominator dominate:

\[ = \lim _{n \rightarrow \infty} \frac{-3n^{3/2}}{2n^{3/2}} = -\frac{3}{2} \]

Final Answer