Why is $\lim_{x \to \infty} x \log(1+\frac{1}{x}) = \lim_{y \to 0} \frac{\log(1+y)}{y}$?
Why is $$\lim_{x \to \infty} x \log(1+\frac{1}{x}) = \lim_{y \to 0} \frac{\log(1+y)}{y} ?$$
I understand that they are both indeterminate forms. Specifically we are initially given $$\lim_{x \to \infty} x \log(1+\frac{1}{x})$$ and have to find the limit. Well with some rewriting we have:
$$\lim_{x \to \infty} x \log(1+\frac{1}{x}) = \\ \lim_{x \to \infty}\frac{x}{\frac{1}{\log(1+\frac{1}{x})}} = \frac{\infty}{\infty} \\\text{(not formally, but I'm just putting it here to stress the point)}$$
Meanwhile
$$\lim_{y \to 0} \frac{\log(1+y)}{y} = \frac{0}{0} \\\text{(by L'Hospital or other arguments the true limit is actually 1, but again to just stress my point)}$$
So they are both indeterminate forms, but they are going to different "limits", what is it about indeterminate forms I'm forgetting to be able to apply ?
$\endgroup$ 13 Answers
$\begingroup$Change the variables $x \mapsto \frac{1}{y}$ and accordingly the limits $(x \to \infty) \mapsto (y \to 0)$
$\endgroup$ 0 $\begingroup$Note that the limit
$$\lim_{y \to 0} \frac{\log(1+y)}{y}=1$$
contains more information since
$$\lim_{y \to 0^+} \frac{\log(1+y)}{y}=\lim_{y \to 0^-} \frac{\log(1+y)}{y}=1$$
indeed we also have
$$\lim_{x \to \infty} x \log\left(1+\frac{1}{x}\right)=\lim_{x \to -\infty} x \log\left(1+\frac{1}{x}\right)=1$$
and both can be derived from the foundamental limit
$$\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x =\lim_{x \to -\infty} \left(1+\frac{1}{x}\right)^x =e$$
$\endgroup$ $\begingroup$It's quite simple: use the substitution $y=\frac1x$, and observe that $\lim_{x\to\infty}y=0$. Therefore, we have$$\lim_{x \to \infty} x \log(1+\frac{1}{x}) = \lim_{y \to 0} \frac{\log(1+y)}{y}$$Next, recall the latter is a high-school limit, which is just the translation from the definition, in terms of limit, that the derivative of $\ln$ at $x=1$ is equal to $1$.
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