Limit of $\frac{t}{\ln(t+1)}$ as $t \to 0$

I need to solve $$\lim_{t\to0} \frac{t}{\ln(1+t)}$$

I know they are equivalent but I can't use my previous knowledge and I am not allowed to solve it with L'Hopital. My instructor only showed that $$\lim_{t\to0} \frac{\sin t}{t} = 1$$

How can I solve it or prove the equivalency? Thanks!

write$$\frac{1}{\frac{1}{t}\ln(1+t)}=\frac{1}{\ln(1+t)^{1/t}}$$ und use that $$\lim_{t\to 0}\left(1+t\right)^{1/t}=e$$

$$\lim_{t\to0} \frac{t}{\ln(1+t)}= \left(\lim_{t\to0} \frac{\ln(1+t)}{t}\right)^{-1} = \left(\lim_{t\to0} \frac{\ln(1+t) -\ln1}{t-0}\right)^{-1} \\=\color{red}{\left(\frac{d}{dt} \ln(1+t)\bigg|_{t =0}\right)^{-1}=1}$$

$$lim_{t\to0}\frac{t}{\ln(t+1)}=lim_{t\to0}(\frac{\ln(t+1)}{t})^{-1}$$

Use the power series for $\ln$:

$$lim_{t\to0}(\frac{\sum_{n=1}^\infty\frac{t^n}{n}}{t})^{-1}=1$$

$\lim_{t\to 0}\frac{\sin t}{t}=1$ implies $\lim_{t\to 0}\frac{1-\cos t}{t^2}=\frac{1}{2}$, since $1-\cos t=2\sin^2\frac{t}{2}$ and $x\to x^2$ is a continuous function. Such limits and De Moivre's formula imply $$\lim_{t\to 0}\frac{e^{it}-1}{t} = i,\qquad \lim_{z\to 0}\frac{e^z-1}{z}=1.$$ By enforcing the substitution $z=\log(1+t)$ in the last limit we get $\lim_{t\to 0}\frac{t}{\log(1+t)}=1$.

• So, yes, all the "classical" remarkable limits turn out to be equivalent to each other. Commented Nov 13, 2017 at 21:13
• And $\lim_{t\to 0}\frac{\sin t}{t}=1$ turns out to be equivalent to the differentiability of the sine function, the rectifiability of the circle, the measurability of the disk etcetera. Commented Nov 13, 2017 at 21:14