Question

Use series to evaluate the limit.\(\mathop {lim}\limits_{x \to 0} \frac{{x - ln(1 + x)}}{{{x^2}}}\)

Short answer

Therefore, The value of the given limit is \(\mathop {lim}\limits_{x \to 0} \frac{{x - ln(1 + x)}}{{{x^2}}} = \frac{1}{2}.\)

Step-by-step solution

Step 1: Given information

The given limit is:

\(\mathop {lim}\limits_{x \to 0} \frac{{x - ln(1 + x)}}{{{x^2}}}\)

Note that the series for\(In(1 + x)\)is given by:

\(\begin{array}{l}In(1 + x) = \sum\limits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\frac{{{x^n}}}{n}} \\ = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + ....\end{array}\)\(\mathop {lim}\limits_{x \to 0} \frac{{x - ln(1 + x)}}{{{x^2}}}\)

Step 2: Substitution the expression into the limits

Substitute\(In(1 + x) = \sum\nolimits_{n = 1}^\infty {{{( - 1)}^{n - 1}}} \frac{{{x^n}}}{n}\)into the limits:\(\mathop {\lim }\limits_{x \to 0} \frac{{x - \left( {\sum\nolimits_{n = 1}^\infty {{{( - 1)}^{n - 1}}\frac{{{x^n}}}{n}} } \right)}}{{{x^2}}}\)

Note that\( - \left( {\sum\nolimits_{n = 1}^\infty {{{( - 1)}^{n - 1}}} \frac{{{x^n}}}{n}} \right)\)can be written as\( + \left( {\sum\nolimits_{n = 1}^\infty {{{( - 1)}^{n - 1}}} \frac{{{x^n}}}{n}} \right)\)by distributing the negative:\(\mathop {\lim }\limits_{x \to 0} \frac{{x + \left( {\sum\nolimits_{n = 1}^\infty {{{( - 1)}^n}\frac{{{x^n}}}{n}} } \right)}}{{{x^2}}}\)

Separate the fractions:

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{{x^2}}} + \frac{{\left( {\sum\nolimits_{n = 1}^\infty {{{( - 1)}^n}\frac{{{x^n}}}{n}} } \right)}}{{{x^2}}}} \right)\)

By the quotient of power property, the fraction \(\frac{{\left( {\sum\nolimits_{n = 1}^\infty {{{( - 1)}^n}\frac{{{x^n}}}{n}} } \right)}}{{{x^2}}}\)can be written as \(\frac{{\left( {\sum\nolimits_{n = 1}^\infty {{{( - 1)}^n}\frac{{{x^{n - 2}}}}{n}} } \right)}}{{{x^2}}}.\)\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{x}{{{x^2}}} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{{x^{n - 2}}}}{n}} } \right)\)

Step 3: Simplify the fraction.

Integrate using the series.

\(\mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} + \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{{x^{n - 2}}}}{n}} } \right)\)

Evaluate some values of the series:

\(\begin{array}{l}\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{{{x^{n - 2}}}}{n}\\{( - 1)^n}\frac{{{x^{n - 2}}}}{n}\end{array}\)

Substitute \(n = 1:\)\({( - 1)^1}\frac{{{x^{1 - 2}}}}{1}\)\( - \frac{1}{x}\)

Step 4: Make a list of the series' terms.

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x}\sum\limits_{n = 1}^\infty {{{( - 1)}^n}\frac{{{x^{n - 2}}}}{n}} } \right)\)

Substitute the limit:

\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{x} - \frac{1}{x} + \frac{1}{2} - \frac{x}{3} + \frac{{{x^2}}}{4} - \frac{{{x^3}}}{5} + \cdots } \right)\)\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{2} - \frac{x}{3} + \frac{{{x^2}}}{4} - \frac{{{x^3}}}{5} + \cdots } \right)\)\( = \frac{1}{2} - 0 + 0 - 0 + \cdots \)\( = \frac{1}{2}\)

The value of the given limit is\(\mathop {lim}\limits_{x \to 0} \frac{{x - ln(1 + x)}}{{{x^2}}} = \frac{1}{2}.\)