Calculate lim x → ∞ 1 ln ⁡ ( x + 1 ) − ln ⁡ ( x ) − x

Question

Calculate       lim          x      →      ∞            1          ln      ⁡      (      x      +      1      )      −      ln      ⁡      (      x      )        −  x

Carmelo Payne · Accepted Answer

If we are allowed to use series expansions, then the problem is fairly trivial. We want to evaluate

lim_{x \to \infty} \frac{1 - x \log (1 + \frac{1}{x})}{\log (1 + \frac{1}{x})}

The series expansion of

\log (1 + \frac{1}{x})

is given by

\log (1 + \frac{1}{x}) = \frac{1}{x} - \frac{1}{2 x^{2}} + \frac{1}{3 x^{3}} - \frac{1}{4 x^{4}} + \dots

Plugging this in, we find that

\frac{1 - x \log (1 + \frac{1}{x})}{\log (1 + \frac{1}{x})} = \frac{1 - x (\frac{1}{x} - \frac{1}{2 x^{2}} + \frac{1}{3 x^{3}} + \dots)}{(\frac{1}{x} - \frac{1}{2 x^{2}} + \frac{1}{3 x^{3}} + \dots)} = \frac{(\frac{1}{2} - \frac{1}{3 x} + \frac{1}{4 x^{2}} \dots)}{(1 - \frac{1}{2 x} + \frac{1}{3 x^{2}} + \dots)} \overset{x \to \infty}{\to} \frac{1}{2}

Thus,

lim_{x \to \infty} \frac{1}{\ln (x + 1) - \ln (x)} - x = \frac{1}{2}

Yesenia Sherman · Accepted Answer

I shall try to evaluate the limit by using L’Hospital Rule twice.

\begin{array}{l} lim_{x \to \infty} (\frac{1}{\ln (x + 1) - \ln x} - x) \\ \begin{aligned} = & lim_{x \to \infty} \frac{1 - x \ln (\frac{x + 1}{x})}{\ln (\frac{x + 1}{x})} (\frac{0}{0}) \\ = & lim_{x \to \infty} \frac{- x (\frac{x}{x + 1}) (- \frac{1}{x^{2}}) - \ln (\frac{x + 1}{x})}{\frac{x}{x + 1} (- \frac{1}{x^{2}})} \\ = & - lim_{x \to \infty} \frac{\frac{1}{x + 1} - \ln (\frac{x + 1}{x})}{\frac{1}{x} - \frac{1}{x + 1}} (\frac{0}{0}) \\ = & - lim_{x \to \infty} \frac{- \frac{1}{(x + 1)^{2}} - \frac{x}{x + 1} (- \frac{1}{x^{2}})}{- \frac{1}{x^{2}} + \frac{1}{(x + 1)^{2}}} \end{aligned} \end{array} \begin{array}{l}  \end{array}

Simplifying the quotient yields

\begin{array}{l} lim_{x \to \infty} (\frac{1}{\ln (x + 1) - \ln x} - x) \\ = lim_{x \to \infty} \frac{\frac{1}{(x + 1)^{2}} - \frac{1}{x (x + 1)}}{- \frac{1}{x^{2}} + \frac{1}{(x + 1)^{2}}} \\ = lim_{x \to \infty} \frac{x^{2} - x (x + 1)}{- (x + 1)^{2} + x^{2}} \\ = lim_{x \to \infty} \frac{- x}{- 2 x + 1} \\ = \frac{- 1}{- 2 + \frac{1}{x}} \\ = \frac{1}{2} \end{array}

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