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Suppose 𝑎𝑛=12𝑛23𝑛+27−4𝑛2+7𝑛+15a n​ = 3n+2712n 2 ​ − n+154n 2 +7​ . Find the value of lim⁡𝑛→∞𝑎𝑛n→∞lim​ a n​ .

Question

Suppose 𝑎𝑛 = 12n23n+274n2+7n+15 \frac{12n^2}{3n + 27 - 4n^2 + 7n + 15}

Find the value of limnan \lim_{n \to \infty} a_n .

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Solution

The given expression is a rational function and the limit as n approaches infinity of a rational function can be found by dividing every term by the highest power of n in the denominator.

The given expression is:

𝑎𝑛 = (12𝑛^2 + 3𝑛 + 27) / (4𝑛^2 + 7𝑛 + 15)

Divide every term by n^2:

𝑎𝑛 = (12 + 3/n + 27/n^2) / (4 + 7/n + 15/n^2)

As n approaches infinity, the terms 3/n, 27/n^2, 7/n, and 15/n^2 will approach 0. So, the expression simplifies to:

𝑎𝑛 = 12 / 4 = 3

Therefore, the value of lim⁡𝑛→∞𝑎𝑛 is 3.

This problem has been solved

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