Question

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

Short answer

The value of \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^{3x}} - {e^{ - 3x}}}}{{{e^{3x}} + {e^{ - 3x}}}}} \right)\) is 1.

Step-by-step solution

Given information

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

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

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

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^{3x}} - {e^{ - 3x}}}}{{{e^{3x}} + {e^{ - 3x}}}}} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\left( {\frac{{{e^{3x}}}}{{{e^{3x}}}} - \frac{{{e^{ - 3x}}}}{{{e^{33}}}}} \right)}}{{\left( {\frac{{{e^{3x}}}}{{{e^{3x}}}} + \frac{{{e^{ - 3x}}}}{{{e^{3x}}}}} \right)}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{1 - {e^{ - 6x}}}}{{\left( {1 + {e^{ - 4x}}} \right)}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{1 - {e^{ - 6x}}}}{{1 + {e^{ - 4x}}}}} \right)\end{array}\)

On further simplification:

\[\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^{3x}} - {e^{ - 3x}}}}{{{e^{3x}} + {e^{ - 3x}}}}} \right) = \frac{{1 - {e^{ - \infty }}}}{{1 + {e^{ - \infty }}}}\\ = \frac{{1 - 0}}{{1 + 0}}\quad \\ = 1\end{array}\]

Therefore, the value of \(\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^{3x}} - {e^{ - 3x}}}}{{{e^{3x}} + {e^{ - 3x}}}}} \right)\) is 1.