The Number e
Solved Problems
Example 5.
Find the limit \[\lim\limits_{x \to \infty } {\left( {\frac{{x + a}}{{x - a}}} \right)^x}.\]
Solution.
We first convert the base of the function:
\[L = \lim\limits_{x \to \infty } {\left( {\frac{{x + a}}{{x - a}}} \right)^x} = \lim\limits_{x \to \infty } {\left( {\frac{{x - a + 2a}}{{x - a}}} \right)^x} = \lim\limits_{x \to \infty } {\left( {1 + \frac{{2a}}{{x - a}}} \right)^x}.\]
Introduce the new variable: \(y = {\frac{{2a}}{{x - a}}}.\) As \(x \to \infty,\) \(y \to 0\) and, hence,
\[x - a = \frac{{2a}}{y},\;\;\; x = a + \frac{{2a}}{y}.\]
Substituting this into the function gives
\[L = \lim\limits_{x \to \infty } {\left( {1 + \frac{{2a}}{{x - a}}} \right)^x} = \lim\limits_{y \to 0} {\left( {1 + y} \right)^{a + \frac{{2a}}{y}}} = \lim\limits_{y \to 0} {\left( {1 + y} \right)^a} \cdot \lim\limits_{y \to 0} {\left( {1 + y} \right)^{\frac{{2a}}{y}}} = 1 \cdot {e^{2a}} = {e^{2a}}.\]
Example 6.
Calculate the limit \[\lim\limits_{x \to \infty } {\left( {\frac{x}{{x + 1}}} \right)^x}.\]
Solution.
First we transform the base:
\[L = \lim\limits_{x \to \infty } {\left( {\frac{x}{{x + 1}}} \right)^x} = \lim\limits_{x \to \infty } {\left( {\frac{{x + 1 - 1}}{{x + 1}}} \right)^x} = \lim\limits_{x \to \infty } {\left( {1 - \frac{1}{{x + 1}}} \right)^x}.\]
Let \({ - {\frac{1}{{x + 1}}}} = y.\) Then
\[x + 1 = - \frac{1}{y},\;\; \Rightarrow x = - \frac{1}{y} - 1\;\;\; \text{and}\;\; y \to 0\;\;\text{as}\;\;x \to \infty .\]
Now we can find the limit:
\[L = \lim\limits_{x \to \infty } {\left( {1 - \frac{1}{{x + 1}}} \right)^x} = \lim\limits_{y \to 0} {\left( {1 + y} \right)^{ - \frac{1}{y} - 1}} = \frac{{\lim\limits_{y \to 0} {{\left( {1 + y} \right)}^{ - \frac{1}{y}}}}}{{\lim\limits_{y \to 0} {{\left( {1 + y} \right)}^{ + 1}}}} = \frac{{\lim\limits_{y \to 0} {{\left[ {{{\left( {1 + y} \right)}^{\frac{1}{y}}}} \right]}^{ - 1}}}}{1} = \left[ {\lim\limits_{y \to 0} {{\left( {1 + y} \right)}^{\frac{1}{y}}}} \right]^{ - 1} = \frac{1}{e}.\]
Example 7.
Evaluate the limit \[\lim\limits_{x \to \infty } {\left( {\frac{{x + 3}}{{x - 2}}} \right)^{x - 1}}.\]
Solution.
We can write this limit as follows:
\[L = \lim\limits_{x \to \infty } {\left( {\frac{{x + 3}}{{x - 2}}} \right)^{x - 1}} = \lim\limits_{x \to \infty } {\left( {\frac{{x - 2 + 5}}{{x - 2}}} \right)^{x - 1}} = \lim\limits_{x \to \infty } {\left( {1 + \frac{5}{{x - 2}}} \right)^{x - 1}} = \lim\limits_{x \to \infty } {\left[ {{{\left( {1 + \frac{5}{{x - 2}}} \right)}^{\frac{{x - 2}}{5}}}} \right]^{\frac{{5\left( {x - 1} \right)}}{{x - 2}}}}.\]
Replace the variable:
\[\frac{5}{{x - 2}} = y,\;\; \Rightarrow x - 2 = \frac{5}{y},\;\; \Rightarrow x = \frac{5}{y} + 2.\]
Here \(y \to 0\) as \(x \to \infty.\) Then the limit is
\[L = \lim\limits_{x \to \infty } {\left[ {{{\left( {1 + \frac{5}{{x - 2}}} \right)}^{\frac{{x - 2}}{5}}}} \right]^{\frac{{5\left( {x - 1} \right)}}{{x - 2}}}} = \lim\limits_{y \to 0} {\left[ {{{\left( {1 + y} \right)}^{\frac{1}{y}}}} \right]^{y\left( {\frac{5}{y} + 2 - 1} \right)}} = \lim\limits_{y \to 0} {\left[ {{{\left( {1 + y} \right)}^{\frac{1}{y}}}} \right]^{5 + y}} = \lim\limits_{y \to 0} {\left[ {{{\left( {1 + y} \right)}^{\frac{1}{y}}}} \right]^5} \cdot \lim\limits_{y \to 0} {\left[ {{{\left( {1 + y} \right)}^{\frac{1}{y}}}} \right]^y} = \lim\limits_{y \to 0} {\left[ {{{\left( {1 + y} \right)}^{\frac{1}{y}}}} \right]^5} \cdot \lim\limits_{y \to 0} \left( {1 + y} \right) = {e^5} \cdot 1 = {e^5}.\]
Example 8.
Find the limit \[\lim\limits_{x \to a} {\frac{{\ln x - \ln a}}{{x - a}}}\;\left( {a \gt 0} \right).\]
Solution.
Let \(x - a = t.\) It is easy to see that \(t \to 0\) as \(x \to a.\) Then
\[L = \lim\limits_{x \to a} \frac{{\ln x - \ln a}}{{x - a}} = \lim\limits_{t \to 0} \frac{{\ln \left( {t + a} \right) - \ln a}}{t} = \lim\limits_{t \to 0} \frac{{\ln \frac{{t + a}}{a}}}{t} = \lim\limits_{t \to 0} \frac{1}{t}\ln \left( {1 + \frac{t}{a}} \right).\]
Make one more change of variable:
\[\frac{t}{a} = z,\;\;z \to 0\;\; \text{as}\;\;t \to 0.\]
Hence, the limit becomes
\[L = \lim\limits_{t \to 0} \frac{1}{t}\ln \left( {1 + \frac{t}{a}} \right) = \lim\limits_{z \to 0} \frac{1}{{az}}\ln \left( {1 + z} \right) = \frac{1}{a}\lim\limits_{z \to 0} \ln {\left( {1 + z} \right)^{\frac{1}{z}}} = \frac{1}{a}\ln \left[ {\lim\limits_{z \to 0} {{\left( {1 + z} \right)}^{\frac{1}{z}}}} \right] = \frac{1}{a}\ln e = \frac{1}{a}.\]
Example 9.
Calculate the limit \[\lim\limits_{x \to 0} {\left( {1 + \sin x} \right)^{\frac{1}{x}}}.\]
Solution.
The limit can be represented in the following form:
\[L = \lim\limits_{x \to 0} {\left( {1 + \sin x} \right)^{\frac{1}{x}}} = \lim\limits_{x \to 0} {\left( {1 + \sin x} \right)^{\frac{1}{{\sin x}} \cdot \frac{{\sin x}}{x}}} = \lim\limits_{x \to 0} {\left[ {{{\left( {1 + \sin x} \right)}^{\frac{1}{{\sin x}}}}} \right]^{\frac{{\sin x}}{x}}}.\]
After taking logarithm, we have
\[\ln L = \ln \left( {\lim\limits_{x \to 0} {{\left[ {{{\left( {1 + \sin x} \right)}^{\frac{1}{{\sin x}}}}} \right]}^{\frac{{\sin x}}{x}}}} \right) = \lim\limits_{x \to 0} \left( {\frac{{\sin x}}{x} \ln \left[ {{{\left( {1 + \sin x} \right)}^{\frac{1}{{\sin x}}}}} \right]} \right) = \lim\limits_{x \to 0} \frac{{\sin x}}{x} \cdot \lim\limits_{x \to 0} \left( {\ln \left[ {{{\left( {1 + \sin x} \right)}^{\frac{1}{{\sin x}}}}} \right]} \right).\]
We notice that \(\lim\limits_{x \to 0} {\frac{{\sin x}}{x}} = 1.\) Besides that, \(\sin x \to 0\) as \(x \to 0,\) therefore, we can replace the transition \(x \to 0\) in the second limit with the equivalent limit \(\sin x \to 0.\) This yields:
\[\ln L = 1 \cdot \lim\limits_{\sin x \to 0} \left( {\ln \left[ {{{\left( {1 + \sin x} \right)}^{\frac{1}{{\sin x}}}}} \right]} \right) = \ln \lim\limits_{\sin x \to 0} \left[ {{{\left( {1 + \sin x} \right)}^{\frac{1}{{\sin x}}}}} \right].\]
As \(\lim\limits_{\sin x \to 0} {{{\left( {1 + \sin x} \right)}^{\frac{1}{{\sin x}}}}} = e,\) we get
\[\ln L = \ln e = 1.\]
Thus, \(L = e.\)
