How to calculate limit $\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$ without using L'Hospital rule

Question

I have to calculate next limit without using L'Hospital rule: $$\lim_{x \rightarrow 0}\frac{x+\ln(e+x)-1}{x}$$ I came this far with that $$1+\frac{1}{e}\lim_{t\rightarrow1}\frac{\ln(t)}{t-1}$$ I know that result for this limit using the L'Hospital rule is $1+\frac{1}{e}$ Any suggestions how to finish this?

Answer 1

If $f(x)=x+\log(x+e)$, then $f(0)=\log(e)=1$. Therefore, your limit is $f'(0)$, which is $1+\frac1e$.

Answer 2

Using your approach, $$\lim_{t\to 1}\frac{\ln t}{t-1}=\lim_{t\to 1}\frac{\ln (1+(t-1))}{t-1}= \lim_{t\to 1}\ln\left(1+(t-1)\right)^{1/(t-1)}$$ Now, recall that $\lim_{x\to 0}(1+x)^{1/x}=e$

Answer 3

So you just need to calculate $ \lim\limits_{x\to 0}{\frac{\ln{\left(1+x\right)}}{x}} $ whithout using L'hopital. We'll not be using series expansion either.

Let $ x\in\left(-1,1\right)\setminus\left\lbrace 0\right\rbrace $. We have : \begin{aligned} \frac{\ln{\left(1+x\right)}}{x}&=\int_{0}^{1}{\frac{\mathrm{d}y}{1+xy}}\\&=1-x\int_{0}^{1}{\frac{y}{1+xy}\,\mathrm{d}y} \end{aligned}

Since : \begin{aligned} \left|x\int_{0}^{1}{\frac{y}{1+xy}\,\mathrm{d}y}\right|\leq\left|x\right|\int_{0}^{1}{\frac{y}{\left|1+xy\right|}\,\mathrm{d}y}&\leq\left|x\right|\int_{0}^{1}{\frac{y}{\left|1-\left|x\right|y\right|}\,\mathrm{d}y}\\&\leq\frac{\left|x\right|}{1-\left|x\right|}\int_{0}^{1}{y\,\mathrm{d}y}=\frac{\left|x\right|}{2\left(1-\left|x\right|\right)}\underset{x\to 0}{\longrightarrow}0 \end{aligned}

Then $ \lim\limits_{x\to 0}{x\int_{0}^{1}{\frac{y}{1+xy}\,\mathrm{d}y}}=0 $, which means : $$ \lim_{x\to 0}{\frac{\ln{\left(1+x\right)}}{x}}=1 $$

Answer 4

Using series expansions in $$y=\frac{x+\log(e+x)-1}{x}$$ $$\log(e+x)=1-\sum_{n=1}^\infty\frac{\left(-\frac{1}{e}\right)^n}{n} x^n=1+\frac{x}{e}-\frac{x^2}{2 e^2}+\frac{x^3}{3 e^3}+O\left(x^4\right)$$ $$y=\left(1+\frac{1}{e}\right)-\frac{x}{2 e^2}+\frac{x^2}{3 e^3}+O\left(x^3\right)$$ Try it with $x=\frac e{10}$ (very far away from $0$). The exact value is $$1+\frac{10 \log \left(\frac{11}{10}\right)}{e}\sim 1.35063$$ while the truncated series gives $$1+\frac{143}{150 e}\sim 1.35071$$