Calculating $\lim_{x \to 0} \frac{1}{x}\ln{\frac{e^{x+1}}{e+x}}$ This is a simple limit problem, but I don't understand why this approach is logically wrong.
$$\lim_{x \to 0} \frac{1}{x}\ln{\frac{e^{x+1}}{e+x}} \\ = \lim_{x \to 0} \frac{\ln{e^{x+1}}-\ln{(e+x)}}{x} \\ = \lim_{x \to 0} \frac{(x+1)\ln{e}-\ln{(e+x)}}{x} \\ = \lim_{x \to 0} \frac{(x+1)-\ln{e}}{x} \\ = \lim_{x \to 0} \frac{(x+1)-1}{x} \\ = 1$$
All I did is switching $\ln{e}$ to $1$, and making $\ln{(e+x)} \to 1$ which seems logical to me, but it gave me wrong answer. What am I doing wrong here?
 A: A simpler example where similar substitution does not work will be $\lim_{x \rightarrow 0} \frac{e-(e+x)}{x}$. Partially plugging $e+x \rightarrow e$ will result in $\lim \frac{0}{x} = 0$, whereas the answer should be $\lim \frac{-x}{x} = -1$.
There are cases when we can do such step. Addition and multiplication works: $\lim_{x \rightarrow h} A(x) + B(x) = a+b$ if $\lim_{x \rightarrow h} A(x) = a < \infty$ and $\lim_{x \rightarrow h} B(x) = b < \infty$ (similarly for $\times$ operation). Division only works when $b \ne 0$ (which in your question, is not). It also works under continuous function chain: $\lim_{x \rightarrow h} A(B(x)) = A(b)$ if $A(x)$ is continuous on $b$.
A: You can't replace $\ln(e+x) $ by $\ln e$
$\begin{align}&\lim_{x \to 0} \frac{1}{x}\ln{\frac{e^{x+1}}{e+x}}\\&= \lim_{x \to 0} \frac{\ln{e^{x+1}}-\ln{(e+x)}}{x}\\&= \lim_{x \to 0} \frac{(x+1)\ln{e}-\ln{(e+x)}}{x} \\&= \lim_{x \to 0} \frac{(x+1)-\ln{(1+\frac{x}{e})}-\ln e}{x} \\&= \lim_{x \to 0} \frac{(x+1)-1}{x}-\lim_{\frac{x}{e}\to 0}\frac{1}{e}\frac{\ln(1+\frac{x}{e})} {\frac{x}{e}}\\&= 1-\frac{1}{e}\end{align}$
Note: $\lim_{x\to 0}\frac{\ln(1+x) }{x}=1$
A: We have:
$$\frac{e^{x+1}}{e+x}=\frac{e\cdot (e^{x}-1)+e}{e+x}\,\,\sim\,\,\frac{ex+e}{e+x}$$
So:
$$\lim_{x\to 0}\frac{1}x\ln\left(\frac{ex+e}{e+x}\right)=\lim_{x\to 0}\frac{1}{x}\ln\left(\frac{e}{e+x}\cdot (1+x)\right)\,\ \sim\,\, \lim_{x\to 0}\frac{1}{x}\ln\left(1-\frac{x}{x+e}\right)+ \lim_{x\to 0}\frac1x \ln \left(1+x\right)=1-\frac{1}{e}$$
A: $$
\frac{1}{x}\log(e^{1+x})-\frac{1}{x}\log(e+x)=\frac{1+x-\log(e+x)}{x}=\frac{1+x-\log(e(1+\frac{x}{e}))}{x}
$$
Now, you can use property of logarithm and you can do a limited developpment of $\log(1+h)$ when $h$ approaches $0$.
$$
\frac{1+x-\log(e(1+\frac{x}{e}))}{x}=\frac{x-\log(1+\frac{x}{e})}{x}=\frac{x-\frac{x}{e}+\frac{x^{2}}{2e^{2}}+O(x^{2})}{x} \xrightarrow[x \rightarrow 0]{}1-\frac{1}{e}
$$
