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I have to calculate some limits and try to solve them in use of taylor.

$$ \lim\limits_{x\to \infty} \left(x-x^2 \ln (1+\frac{1}{x})\right) $$

In taylor pattern I have $x_0$ to put, but there $x_0$ is $\infty$ so I want to replace it with something other $$ y = \frac{1}{x} \\ \lim_{y\to 0^+} \left(\frac{1}{y}-\frac{1}{y^2} \ln (1+y)\right) $$

Let $$ f(y) = \frac{1}{y}-\frac{1}{y^2} \ln (1+y) $$ $$f'(y) = -\frac{1}{y^2} + \left(-\frac{2}{y^3}\ln (1+y) - \frac{y^2}{1+y}\right) $$

but $f'(0)$ does not exists because I have $0$ in denominator.

MP3129
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  • What are you trying to achieve with the derivative at $0$. If your function is the square root, your derivative will be infinity at $0$. However its limit does exist. – Stan Tendijck Mar 07 '19 at 14:25
  • If Taylor's expansion is not mandatory, see https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%C3%B4pital-rule-or-series-expansion – lab bhattacharjee Mar 07 '19 at 14:41

2 Answers2

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You just have to write the Taylor expansion of $\ln \left( 1 + \frac{1}{x} \right)$ when $x$ tends to $+\infty$ : $$x - x^2 \ln \left( 1 + \frac{1}{x} \right) = x - x^2 \left( \frac{1}{x} - \frac{1}{2x^2} + o\left( \frac{1}{x^2} \right) \right) = \frac{1}{2} + o(1)$$

So the limit is equal to $\frac{1}{2}$.

TheSilverDoe
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Proceeding with your substitution, since $\log \left( 1+y\right) =y-\frac{1}{ 2}y^{2}+O\left( y^{3}\right) $, we have:

\begin{eqnarray*} \frac{1}{y}-\frac{1}{y^{2}}\log \left( 1+y\right) &=&\frac{1}{y}-\frac{1}{ y^{2}}\left( y-\frac{1}{2}y^{2}+O\left( y^{3}\right) \right) &=&\frac{1}{2}+O\left( y\right) \overset{y\rightarrow 0}{\longrightarrow } \frac{1}{2}. \end{eqnarray*}

Américo Tavares
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