Question

To determine the value of \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {{e^{\frac{3}{{2 - x}}}}} \right)\).

Short answer

The value of \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {{e^{\frac{3}{{2 - x}}}}} \right)\) is \(\infty \).

Step-by-step solution

Given information

The expression is \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {{e^{\frac{3}{{2 - x}}}}} \right)\).

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 \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {{e^{\frac{3}{{2 - x}}}}} \right)\)

The value of\(\mathop {\lim }\limits_{x \to {2^ - }} \left( {{e^{\frac{3}{{2 - x}}}}} \right)\)is shown below.

\(\mathop {\lim }\limits_{x \to {2^ - }} \left( {{e^{\frac{3}{{2 - x}}}}} \right) \to \infty \)

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

So, the\(\mathop {\lim }\limits_{x \to {2^ - }} \left( {{e^{\frac{3}{{2 - x}}}}} \right)\)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:

\(\begin{array}{c}\mathop {\lim }\limits_{x \to {2^ - }} \left( {{e^{\frac{3}{{2 - x}}}}} \right) = \mathop {\lim }\limits_{t \to \infty } \left( {{e^t}} \right)\\ = {e^\infty }\\ = \infty \end{array}\)

Therefore, the value of \(\mathop {\lim }\limits_{x \to {2^ - }} \left( {{e^{\frac{3}{{2 - x}}}}} \right)\) is \(\infty \).