Question

Find the limit or show that it does not exist.

42. \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {{\rm{ln}}\left( {2 + x} \right) - {\rm{ln}}\left( {1 + x} \right)} \right)\)

Short answer

The value of the limit is \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {{\rm{ln}}\left( {2 + x} \right) - {\rm{ln}}\left( {1 + x} \right)} \right) = 0\).

Step-by-step solution

Apply logarithm property

Apply the logarithmic property \({\rm{ln}}a - {\rm{ln}}b = {\rm{ln}}\frac{a}{b}\).

Use the property and simplify the limit expression.

\(\begin{aligned}\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {{\rm{ln}}\left( {2 + x} \right) - {\rm{ln}}\left( {1 + x} \right)} \right) &= \mathop {{\rm{lim}}}\limits_{x \to \infty } {\rm{ln}}\frac{{\left( {2 + x} \right)}}{{1 + x}}\\ &= {\rm{ln}}\left( {\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{\left( {2 + x} \right)}}{{1 + x}}} \right)\\ &= {\rm{ln}}\left( {\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{\left( {\frac{2}{x} + 1} \right)}}{{\frac{1}{x} + 1}}} \right)\end{aligned}\)

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

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

Apply sum laws

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

\[\begin{aligned}{\rm{ln}}\left( {\frac{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {\frac{2}{x} + 1} \right)}}{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {\frac{1}{x} + 1} \right)}}} \right) &= {\rm{ln}}\left( {\frac{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{2}{x} + \mathop {{\rm{lim}}}\limits_{x \to \infty } 1}}{{\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{1}{x} + \mathop {{\rm{lim}}}\limits_{x \to \infty } 1}}} \right)\\ &= {\rm{ln}}\left( {\frac{{0 + 1}}{{0 + 1}}} \right)\\ &= {\rm{ln}}\left( 1 \right)\\ &= 0\end{aligned}\]

Thus, the value of the limit is \(0\).