Limit of $\frac{\log(e+x)-1}{x}$ as $x\to 0$ without using L'Hopital

Question

I was doing a derivative by definition and I need to solve that limit using equivalent infinitesimals and such, without L'Hopital Rule. Any hint?

Answer 1

Hint: $$\lim_{x\to 0}\frac{\log(e+x)-1}{x}=\lim_{x\to 0}\frac{\log(e+x)-\ln e}{x}=\frac{1}{e}\lim_{x\to 0}\frac{\log(1+\frac{x}{e})}{\frac{x}{e}}$$ and use equivalency $x \sim \ln(1+x)$

Answer 2

$$\lim_{x\to0}\frac{\log e(1+\frac xe) -1}x\\=\lim_{x\to0}\frac{\log e+\log(1+\frac xe) -1}x\\=\lim_{x\to0}\frac {\log(1+\frac xe) }x\\=\lim_{x\to0}\frac{\frac xe-\frac{(\frac xe)^ 2}2+...}x\\=\frac 1e$$

Answer 3

We can define $\ln x=\int_1^x(1/t)dt$ for $x>0.$ With the substitution $u=At$ we have $$\ln A+\ln B=\int_1^A(1/t)dt +\int_1^B(1/u)du=$$ $$=\int_1^A(1/t)dt+\int_{t=A}^{AB}(1/At)dAt=$$ $$=\int_1^A(1/t)dt+\int_A^{AB}(1/t)dt=$$ $$=\int_1^{AB}(1/t)dt=\ln AB.$$

For your Q, we have $-1+\ln (e+x)=-1+\ln e(1+x/e)=$ $=-1+1+\ln (1+x/e)=\ln (1+x/e).$

Observe that if $1+x/e> 0$ then $$\frac {x/e}{1+x/e}=\int_1^{1+x/e}\frac {1}{1+x/e}dt\le $$ $$\le \int_1^{1+x/e}(1/t)dt=\ln (1+x/e)\le $$ $$\le \int_1^{1+x/e}1\cdot dt=x/e.$$