# 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 ?

• Substitute $y= 1/x$. Sep 2, 2020 at 18:23

Change the variables $$x \mapsto \frac{1}{y}$$ and accordingly the limits $$(x \to \infty) \mapsto (y \to 0)$$

Note that the limit

$$\lim_{y \to 0} \frac{\log(1+y)}{y}=1$$

$$\lim_{y \to 0^+} \frac{\log(1+y)}{y}=\lim_{y \to 0^-} \frac{\log(1+y)}{y}=1$$
$$\lim_{x \to \infty} x \log\left(1+\frac{1}{x}\right)=\lim_{x \to -\infty} x \log\left(1+\frac{1}{x}\right)=1$$
$$\lim_{x \to \infty} \left(1+\frac{1}{x}\right)^x =\lim_{x \to -\infty} \left(1+\frac{1}{x}\right)^x =e$$
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 $$1$$.