Final answer:
The limit of the function tan⁻¹(3-x)/(1+√3x) as x approaches infinity is actually 0, not - π/2, because the numerator approaches a constant whereas the denominator increases without bound, causing the overall fraction to approach zero.
Step-by-step explanation:
We are tasked to calculate the limit of the function tan⁻¹(3-x)/(1+√3x) as x approaches infinity. Let's break it down step by step:
Firstly, we notice that as x becomes very large, the term 3-x becomes more and more negative, and thus tan⁻¹(3-x) approaches - π/2 because the arctangent of a large negative number is close to - π/2. On the other hand, the denominator 1+√3x grows without bound as x becomes large.
As a result, the overall fraction tan⁻¹(3-x)/(1+√3x) approaches zero because the numerator approaches a constant value (- π/2), while the denominator increases without limit, leading the whole fraction towards zero.
Therefore, the limit of the function as x approaches infinity is 0, not - π/2 as was suggested. It's a common mistake to mix up the behavior of the arctangent function as its argument approaches negative infinity with the limit of the entire fraction.
Therefore, the limit of the function tan⁻¹(3-x)/(1+√3x) as x approaches infinity is actually 0, not - π/2.