Question

To determine the value of \(\mathop {\lim }\limits_{x \to m} \tanh x\) by using definitions of hyperbolic functions.

Short answer

The value of \(\mathop {\lim }\limits_{x \to \infty } \tanh x\) is 1.

Step-by-step solution

Given data

The given expression is \(\mathop {\lim }\limits_{x \to \infty } \tanh x\)

Concept used of hyperbolic function

Hyperbolic function:

\(\begin{array}{l}\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2},\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2},\tanh x = \frac{{\sinh x}}{{\cosh x}},\\{\mathop{\rm csch}\nolimits} x = \frac{1}{{\sinh x}},{\mathop{\rm sech}\nolimits} x = \frac{1}{{\cosh x}},\coth x = \frac{{\cosh x}}{{\sinh x}}\end{array}\)

Evaluate the limit \(\mathop {\lim }\limits_{x \to m} \tanh x\) 

The given expression is\(\mathop {\lim }\limits_{x \to \infty } \tanh x\).

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \tanh x = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\sinh x}}{{\cosh x}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}} \right)\end{array}\)

Multiplying numerator and denominator by\({e^{ - x}}\).

\(\begin{array}{c} = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}} \cdot \frac{{{e^{ - x}}}}{{{e^{ - x}}}}} \right)\quad \\ = \mathop {\lim }\limits_{x \to w} \left( {\frac{{{e^x}{e^{ - x}} - {e^{ - x}}{e^{ - x}}}}{{{e^x}{e^{ - x}} + {e^{ - x}}{e^{ - x}}}}} \right)\end{array}\)

Put the value of \(x = \infty \)

Simplify further as shown below.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \tanh x = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{1 - {e^{ - 2x}}}}{{1 + {e^{ - 2x}}}}} \right)\\ = \frac{{1 - {e^{ - x}}}}{{1 + {e^{ - x}}}}\\ = \frac{{1 - 0}}{{1 + 0}}\end{array}\)

Since,\({e^{ - \infty }} = 0\).

\( = 1\)

Therefore, the value of \(\mathop {\lim }\limits_{x \to \infty } \tanh x\) is 1.