Question

15–42: Find the limit or show that it does not exist.

38. \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{{e^{3x}} - {e^{ - 3x}}}}{{{e^{3x}} + {e^{ - 3x}}}}\)

Short answer

The value of the limit is \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{{e^{3x}} - {e^{ - 3x}}}}{{{e^{3x}} + {e^{ - 3x}}}} = 1\).

Step-by-step solution

Simplify the limit expression

Divide the numerator and the denominator by \({e^{3x}}\).

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

Apply Quotient laws

According to the Quotient law, \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).

\(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{1 - {e^{ - 6x}}}}{{1 + {e^{ - 6x}}}} = \frac{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {1 - {e^{ - 6x}}} \right)}}{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {1 + {e^{ - 6x}}} \right)}}\)

Apply Difference and Sum laws

According to the Difference and sum law, \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) - \mathop {\lim }\limits_{x \to a} g\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) + g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) + \mathop {\lim }\limits_{x \to a} g\left( x \right)\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists.

\(\begin{array}{c}\frac{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {1 - {e^{ - 6x}}} \right)}}{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {1 + {e^{ - 6x}}} \right)}} = \frac{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( 1 \right) - \mathop {{\rm{lim}}}\limits_{x \to \infty } {e^{ - 6x}}}}{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( 1 \right) + \mathop {{\rm{lim}}}\limits_{x \to \infty } {e^{ - 6x}}}}\\ = \frac{{1 - 0}}{{1 + 0}}\\ = 1\end{array}\)

Thus, the value of the limit is \(1\).