Answer :
To find the limit of the function [tex]\( \lim_{x \rightarrow \infty} \tan^{-1}\left(\frac{1 - \sqrt{3} x}{3 + x}\right) \)[/tex], we'll follow these steps:
1. Identify the behavior of the numerator and denominator as [tex]\( x \)[/tex] approaches infinity.
The given function inside the arctan function is [tex]\( \frac{1 - \sqrt{3} x}{3 + x} \)[/tex].
- As [tex]\( x \to \infty \)[/tex]:
- The numerator [tex]\( 1 - \sqrt{3} x \)[/tex] will be dominated by the term [tex]\(-\sqrt{3} x\)[/tex], which approaches [tex]\(-\infty\)[/tex].
- The denominator [tex]\( 3 + x \)[/tex] will be dominated by the term [tex]\( x \)[/tex], which approaches [tex]\( \infty \)[/tex].
2. Analyze the behavior of the fraction:
- As [tex]\( x \to \infty \)[/tex], the fraction [tex]\( \frac{1 - \sqrt{3} x}{3 + x} \approx \frac{-\sqrt{3} x}{x} = -\sqrt{3} \)[/tex].
3. Evaluate the function:
Since the fraction [tex]\( \frac{1 - \sqrt{3} x}{3 + x} \)[/tex] approaches [tex]\( -\sqrt{3} \)[/tex] as [tex]\( x \to \infty \)[/tex], we need to find the arctan of this limit:
- We know that as [tex]\( x \to \infty \)[/tex], the fraction approaches a constant value [tex]\( -\sqrt{3} \)[/tex].
Therefore, we evaluate the limit:
[tex]\[ \lim_{x \rightarrow \infty} \tan^{-1}\left(\frac{1 - \sqrt{3} x}{3 + x}\right) = \tan^{-1}(-\sqrt{3}) \][/tex]
4. Using the properties of the arctan function:
We know that:
[tex]\[ \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3} \][/tex]
So, taking the arctan of [tex]\(-\sqrt{3}\)[/tex] will give us:
[tex]\[ \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \][/tex]
Thus, the final answer is:
[tex]\[ \boxed{-\frac{\pi}{3}} \][/tex]
1. Identify the behavior of the numerator and denominator as [tex]\( x \)[/tex] approaches infinity.
The given function inside the arctan function is [tex]\( \frac{1 - \sqrt{3} x}{3 + x} \)[/tex].
- As [tex]\( x \to \infty \)[/tex]:
- The numerator [tex]\( 1 - \sqrt{3} x \)[/tex] will be dominated by the term [tex]\(-\sqrt{3} x\)[/tex], which approaches [tex]\(-\infty\)[/tex].
- The denominator [tex]\( 3 + x \)[/tex] will be dominated by the term [tex]\( x \)[/tex], which approaches [tex]\( \infty \)[/tex].
2. Analyze the behavior of the fraction:
- As [tex]\( x \to \infty \)[/tex], the fraction [tex]\( \frac{1 - \sqrt{3} x}{3 + x} \approx \frac{-\sqrt{3} x}{x} = -\sqrt{3} \)[/tex].
3. Evaluate the function:
Since the fraction [tex]\( \frac{1 - \sqrt{3} x}{3 + x} \)[/tex] approaches [tex]\( -\sqrt{3} \)[/tex] as [tex]\( x \to \infty \)[/tex], we need to find the arctan of this limit:
- We know that as [tex]\( x \to \infty \)[/tex], the fraction approaches a constant value [tex]\( -\sqrt{3} \)[/tex].
Therefore, we evaluate the limit:
[tex]\[ \lim_{x \rightarrow \infty} \tan^{-1}\left(\frac{1 - \sqrt{3} x}{3 + x}\right) = \tan^{-1}(-\sqrt{3}) \][/tex]
4. Using the properties of the arctan function:
We know that:
[tex]\[ \tan\left(-\frac{\pi}{3}\right) = -\sqrt{3} \][/tex]
So, taking the arctan of [tex]\(-\sqrt{3}\)[/tex] will give us:
[tex]\[ \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \][/tex]
Thus, the final answer is:
[tex]\[ \boxed{-\frac{\pi}{3}} \][/tex]
