find the limit.\n lim_{x\rightarrow0^{+}}\tan^{-1}left(\frac{2}{x}\right)
find the limit.\n lim_{x\rightarrow0^{+}}\tan^{-1}left(\frac{2}{x}\right)
Answer
Explanation:
Step1: Analyze the behavior of $\frac{2}{x}$ as $x\to0^{+}$
As $x\to0^{+}$, $\frac{2}{x}\to+\infty$.
Step2: Recall the limit of inverse - tangent function
We know that $\lim_{u\to+\infty}\tan^{- 1}(u)=\frac{\pi}{2}$. Let $u = \frac{2}{x}$. When $x\to0^{+}$, $u\to+\infty$. So $\lim_{x\to0^{+}}\tan^{-1}(\frac{2}{x})=\lim_{u\to+\infty}\tan^{-1}(u)$.
Answer:
$\frac{\pi}{2}$