Find $\lim _{x\to \infty }\left(x\left(\ln\left(x+1\right)-\ln x\right)\right) $

Question

Find

$$\lim _{x\to \infty }x\Big(\ln\left(x+1\right)-\ln x\Big)$$

Here's how I do it:

$$x\Big(\ln(x+1)-\ln x\Big) = x\Bigg(\ln(x(1+\frac{1}{x}) - \ln x\Bigg)$$ $$x\Big(\ln x + \ln(1+\frac{1}{x}\Big)-\ln x =x\ln\Big(1+\frac{1}{x}\Big)$$ $$\ln\Bigg( \Big(1+\frac{1}{x}\Big)^x \Bigg) \rightarrow \ln 1^\infty = 0$$

What am I doing wrong? The answer is supposed to be $1$, but I get $0$.

Answer 1

Just write it as

$$ \lim_{x \to \infty} x\ln(1+1/x) = \lim_{t \to 0} \frac{ \ln(1+t)}{t}=\dots. $$

I think you can finish it!

Answer 2

$1^\infty\neq 1$. It is a case you have to investigate. Here, $$ \lim_{x\to +\infty}\bigl(1+1/x\bigr)^x = e. $$ This should be derived in your book.

Answer 3

Hint: Let $y = \frac{1}{x}$ then $$\lim_{y\to 0} \frac{\ln( 1 + y)}{y} = \ldots$$

Can you get it from here?

Note: $\lim_{x\to \infty} (1 + \frac{1}{x})^x = e$ and $\ln e = 1$.

Answer 4

look at $$x\left(\ln(x+1) - \ln(x) \right) = x \ln\left(\dfrac{x+1}{x}\right) = x\left( \ln(1 + 1/x) \right) = x\left(1/x -1/(2x^2) + \cdots \right) = 1$$ for $x \to \infty$