I need help with the following limit:
lim
I tried using \ln, taking out the n, writing it as \frac{1}{n} in the denominator and performing l'Hôpital (and then using the exponent to get the limit). But for some reason I'm not reaching the correct limit. I would greatly appreciate if someone could show me steps for solving this limit. Thank you.
Answer
Hint:
\begin{align} \lim_{n \to \infty} n\big(n \ln(1+\frac{1}{n})-1\big) & = \lim_{x \to 0^+} \frac{1}{x}\big(\frac{1}{x} \ln(1+x)-1\big) = \\ &= \lim_{x \to 0^+} \frac{\ln(1+x)-x}{x^2} \\ &= \lim_{x \to 0^+} \frac{\cfrac{1}{1+x}-1}{2 x} = \cdots \\ \end{align}
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