Question

To determine the value of \(\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} \left( {{e^{\tan x}}} \right).\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} \left( {{e^{\tan x}}} \right)\) is .

Step-by-step solution

Given information

The expression \(\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} \left( {{e^{\tan x}}} \right).\)

Step 2: Concept of limits

Limit, mathematical concept based on the idea of closeness, used primarily to assign values to certain functions at points where no values are defined, in such a way as to be consistent with nearby values.

Obtain the value of \(\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} \left( {{e^{\tan x}}} \right)\)

Let,\(t = \tan x\), compute\(\mathop {\lim }\limits_{x \to {{\left( {\frac{x}{2}} \right)}^ + }} (\tan x)\).

\(\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} (\tan x) = - \infty \)

Thus, it can be stated that\(t \to - \infty \)as\(x \to {\left( {\frac{\pi }{2}} \right)^ + }\).

So, the\(\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} \left( {{e^{\tan x}}} \right)\)can be written as\(\mathop {\lim }\limits_{t \to - \infty } \left( {{e^t}} \right)\).

Compute\(\mathop {\lim }\limits_{t \to - \infty } \left( {{e^t}} \right)\)as follows:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} \left( {{e^{\tan x}}} \right) = \mathop {\lim }\limits_{t \to - \infty } \left( {{e^t}} \right)\\ = {e^{ - \infty }}\\ = 0\end{array}\)

Therefore, the value of \(\mathop {\lim }\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ + }} \left( {{e^{\tan x}}} \right)\) is