Why is $\lim_{x \to \infty} x \log(1+\frac{1}{x}) = \lim_{y \to 0} \frac{\log(1+y)}{y}$

calculusindeterminate-formsreal-analysis

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 ?

Best Answer

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

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