Why $\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}=1$? I have $\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}=1$
For $x \to 3$, I get: $\frac{0}{0}$
How to calculate it, without L'Hôpital's rule?
 A: With $3-x=1/n$,
$$\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}=\lim_{n\to\infty}\frac1{n\ln\left(1+\dfrac1n\right)}=\lim_{n\to\infty}\frac1{\ln\left(\left(1+\dfrac1n\right)^n\right)}.$$
You should be able to conclude.
A: Let $y = 3-x$, and use the well-known property of limit about $\log$ that:
$\displaystyle \lim_{ y \to 0} \dfrac{\log(y+1)}{y} = 1$ to get the answer. This can be proven by the squeeze theorem. Note that if $x > 0$, then $\dfrac{x}{1+x} < \ln(x+1) < x$. From this we deduce the limit. We can consider the case $x < 0$ similarly.
A: In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities 

$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$

for $x>0$.  
Using $(1)$ we have
$$1=\frac{3-x}{3-x}\le \frac{3-x}{\log(4-x)}\le \frac{3-x}{\frac{3-x}{4-x}}=4-x \tag 2$$
whereupon applying the squeeze theorem yields the coveted limit

$$\lim_{x\to 3}\frac{3-x}{\log(4-x)}=1$$

A: $$\lim\limits_{x \to 3} \frac{3-x}{\ln(4-x)}$$
Let $t = 3 - x$
$$\lim\limits_{t \to 0} \frac{t}{\ln(1+ t)} = \lim\limits_{t \to 0} \frac{t/t}{\ln(1+ t)/t} = 1$$
Proof :-
$\displaystyle \ln(x + 1)/x = y \implies 1+x = e^{xy} \implies {(e^{xy} - 1)y\over xy} = 1 \implies \lim_{xy \to 0 } {(e^{xy} - 1)y\over x} \times \lim_{x \to 0}y = 1 \implies \lim_{x \to 0}y = 1$
