Question

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

37. \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{1 - {e^x}}}{{1 + 2{e^x}}}\)

Short answer

The limit is \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{1 - {e^x}}}{{1 + 2{e^x}}} = - \frac{1}{2}\).

Step-by-step solution

Simplify the limit expression

Divide the numerator and denominator by \({e^x}\) as shown below:

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

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\(\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).

\(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{\frac{1}{{{e^x}}} - 1}}{{\frac{1}{{{e^x}}} + 2}} = \frac{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {\frac{1}{{{e^x}}} - 1} \right)}}{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {\frac{1}{{{e^x}}} + 2} \right)}}\)

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.

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

Thus, the value of the limit is \( - \frac{1}{2}\).