Chapter 3.6, Problem 40E

To determine

To calculate: The value oftheexpression $\underset{x\to 0^{+}}{\lim }{\tan }^{-1}\left(\ln x\right)$ .

Answer to Problem 40E

The value of trigonometric expression $\underset{x\to 0^{+}}{\lim }{\tan }^{-1}\left(\ln x\right)$ is $-\frac{\pi }{2}$ .

Explanation of Solution

Given information:

The expression $\underset{x\to 0^{+}}{\lim }{\tan }^{-1}\left(\ln x\right)$ .

Formula used:

Inverse tangent trigonometric function is expressed as ${\tan }^{-1}\left(x\right)=y$ which is equivalent to $x=\tan y$ where $-\frac{\pi }{2}<y<\frac{\pi }{2}$ .

Domain of inversetangent trigonometric function is $\left(-\infty ,\infty \right)$ and range is $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ .

Calculation:

Consider the expression $\underset{x\to 0^{+}}{\lim }{\tan }^{-1}\left(\ln x\right)$ .

Observe from the above expression that x approaches 0 from right hand side on coordinate axes. So, the logarithmic function tends to negative infinity as x approaches 0 from right hand side.

Now, in a right angled triangle an angle must be there which gives a very large negative value for opposite of side triangle divided by the adjacent side of the triangle.

This is only possible when one of the angles in a right angled triangle approaches $\frac{\pi }{2}$ .

Recall that the inverse tangent trigonometric function is expressed as ${\tan }^{-1}\left(x\right)=y$ which is equivalent to $x=\tan y$ where $-\frac{\pi }{2}<y<\frac{\pi }{2}$ .

Recall that value of tangent function at an angle $\frac{\pi }{2}$ is $\infty$ .

That is, $\tan \left(\frac{\pi }{2}\right)=\infty$ .

Therefore,

  $\begin{array}{c}\underset{x\to 0^{+}}{\lim }{\tan }^{-1}\left(\ln x\right)={\tan }^{-1}\left(\ln 0\right)\\ ={\tan }^{-1}\left(-\infty \right)\\ =-\frac{\pi }{2}\end{array}$

Thus, the value of trigonometric expression $\underset{x\to 0^{+}}{\lim }{\tan }^{-1}\left(\ln x\right)$ is $-\frac{\pi }{2}$ .