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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!

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4 Answers 4

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

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

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

Use the power series for $\ln$:

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

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$\lim_{t\to 0}\frac{\sin t}{t}=1$ 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 $\lim_{t\to 0}\frac{t}{\log(1+t)}=1$.

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  • $\begingroup$ So, yes, all the "classical" remarkable limits turn out to be equivalent to each other. $\endgroup$ Commented Nov 13, 2017 at 21:13
  • $\begingroup$ 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. $\endgroup$ Commented Nov 13, 2017 at 21:14

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