Evaluate $\lim_{x\rightarrow \infty}x\cdot \ln(1+\frac{1}{x})$ without l'Hôpital's Rule

Question

This seems relatively easy to do with l'Hôpital, once you establish the following equality:

$x\cdot \ln(1+\frac{1}{x}) = \frac{\ln(1+\frac1x) }{\frac{1}{x}}$

Take the derivatives and the answer is 1.

What I'm trying to do is to solve it without using l'Hôpital, but I'm quite stuck. Any help would be appreciated, thanks.

Answer 1

You could use $$y-\frac12y^2\le \ln(1+y)<y$$ for $y>0$. This is from using the Taylor series as an alternating series.

Or use the mean value theorem for the fraction: $$ \frac{\ln(1+y)-\ln(1)}{(1+y)-1} $$

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Depending on if that was your definition of the logarithm, you could also use $$ x\ln(1+\frac1x)=x(\ln(x+1)-\ln(x))=x\int_x^{x+1}\frac{dt}t=\int_0^1\frac{x\,ds}{x+s}=\int_0^1\frac{ds}{1+\frac{s}x} $$ and as the integrand is bounded and continuous and the integration interval compact, you can switch limit and integral to get $\int_0^1ds=1$ as result.

Answer 2

Let $g(u)=\ln(1+u)$.

$$\lim_{u\to 0}\frac{\ln(1+u)}{u}=\lim_{u\to 0}\frac{\ln(u+1)-\ln(1)}{u-0}=\lim_{u\to 0}\frac{g(u)-g(0)}{u-0}=g'(0)=1.$$

Answer 3

You might want to take a look at this novel one page proof: http://aleph0.clarku.edu/~djoyce/ma122/elimit.pdf which explains without l'Hôpital's rule that $$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n=e.$$ Now your question just involves taking the natural logarithm of both sides from above. So your answer follows from this quite easily, and is given by $$\ln e=1.$$