Solve $\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x}$ without using L'Hopital or Taylor Series

Question

Solve $$\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x}$$ without using L'Hopital or Taylor Series

I tried:

\begin{align}\lim_{x \rightarrow \infty}\frac{\ln(x+1)-\ln(x)}{x} &= \frac{\ln(x+1)}{x}-\frac{\ln(x)}{x} \\ &=\lim \frac{\ln(x+1)}{x}-\lim\frac{\ln(x)}{x} \\ &=\lim \frac{\ln(x+1)}{x}-0 \\ &=\lim \frac{\ln(x+1)}{x} \\ &=\lim_{x \rightarrow 0} \frac{\ln\left(\frac{1}{x}+1\right)}{\frac{1}{x}} \\ &=\lim_{x \rightarrow 0} x\ln\left(\frac{1}{x}+1\right) \\ &=\lim_{x \rightarrow 0} \ln\left(\left(1+\frac{1}{x}\right)^x\right) \\ &= ???\end{align}

How do I solve this? What do I do next?

Answer 1

The denominator clearly $\to\infty$

Now for the numerator using $\log a/b=\log a-\log b$ for all the logarithm remain defined,

$$\lim_{x\to\infty}(\ln(1+x)-\ln x)=\lim_{x\to\infty}\ln\left(1+\dfrac1x\right)=?$$

Alternatively,

Set $1/x=h$ to get $$\lim_{h\to0}h\cdot\ln(1+h)$$

Answer 2

Recall that $$ \ln x := \int_1^x \frac{1}{t} ~dt.$$

Thus

\begin{align} \lim_{x \to \infty} \frac{\ln (x+1) - \ln x}{x} &= \lim_{x \to \infty} \frac{1}{x} \left[ \int_1^{x+1} \frac{1}{t} ~dt - \int_1^x \frac{1}{t} ~dt \right] \\ &= \lim_{x \to \infty} \frac{1}{x} \int_x^{x+1} \frac{1}{t} ~dt \end{align}

Answer 3

$$\lim_{x\to +\infty}\frac{\log(x+1)-\log(x)}{x}=\lim_{y\to +\infty}\frac{\log(1+e^y)-y}{e^y}=\lim_{y\to +\infty}\frac{\log(1+e^{-y})}{e^y}=0$$ by squeezing: for any $x>0$ we have $\log(1+x)\leq x$ and $e^x\geq x+1$.

Answer 4

Let $f(x)=\ln x$. Now use Lagrange's MVT in $[x,x+1]$ to get $$ f(1+x)-f(x)=f'(\xi), \xi\in(x,1+x) $$ or $$ \ln(1+x)-\ln x=\frac1{\xi}.$$ So $$ \lim_{x\to\infty}\frac1x\ln\frac{1+x}x=0.$$

Answer 5

For $x>1,$

$$0 <\frac{\ln(x+1) - \ln x}{x} < \frac{\ln(x+1)}{x} < \frac{\ln (2x)}{x} = \frac{\ln 2 + \ln x}{x}.$$

Because $(\ln x)/x \to 0$ (a result you appear to know), the limit is $0$ by the squeeze theorem.