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In (1.13.4), divide both sides by λ ᵏWhat can you conclude about (λ2​ /λ 1⁾ᵏas k→[infinity] ? Why can you make similar conclusions about (λ j​ /λ 1​ ⁾ᵏ for j=3…n ?



1 Answer

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Final answer:

As *k* → ∞, *λ*₂ᵏ / *λ*₁ᵏ approaches zero in (1.13.4). Similar conclusions can be drawn for (*λ*ₖ / *λ*₁)ᵏ for *j*=3 to *n*.

Step-by-step explanation:

In (1.13.4), when both sides are divided by *λ*ᵏ, it yields *λ*₂ᵏ / *λ*₁ᵏ = (*λ*₂ / *λ*₁)ᵏ. As *k* → ∞, the ratio *λ*₂ / *λ*₁ raised to the power of *k* tends towards zero since *λ*₂ / *λ*₁ is less than 1. This implies that *λ*₂ᵏ / *λ*₁ᵏ approaches zero as *k* becomes very large.

Similar conclusions can be made for (*λ*ₖ / *λ*₁)ᵏ for *j*=3 to *n*. If *λ*ₖ / *λ*₁ is less than 1, then as *k* → ∞, (*λ*ₖ / *λ*₁)ᵏ will approach zero. This behavior is because the denominator *λ*₁ᵏ grows faster than the numerator *λ*ₖᵏ as *k* becomes larger, resulting in the entire fraction approaching zero.

In summary, dividing both sides of (1.13.4) by *λ*ᵏ allows us to observe the behavior of the ratio *λ*₂ / *λ*₁ as *k* approaches infinity. Similar conclusions can be drawn for the ratios (*λ*ₖ / *λ*₁) for *j*=3 to *n*, enabling insights into the behavior of these terms in the limit.

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