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I know that: $$\lim_{a\to 0} \frac{\ln(1+a)}{a}=1$$

But why $$\lim_{x\to 0} \frac{-\frac{2x}{e}}{\ln(1+(-\frac{2x}{e}))}$$ is 1 too?

$$\lim_{x\to 0} 1 : \frac{\ln(1+(-\frac{2x}{e}))}{-\frac{2x}{e}}=1:\lim_{x\to 0} \frac{\ln(1+(-\frac{2x}{e}))}{-\frac{2x}{e}}=1:1=1$$

It's right?

Dave
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3 Answers3

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Let $y=-2x/e$. Then, as $x\to 0$, $y\to 0$.

Next, we see that

$$\frac{(-2x/e)}{\log(1+(-2x/e))}=\frac{y}{\log(1+y)}=\frac{1}{\frac{\log(1+y)}{y}}$$

Hence, we can assert that

$$\lim_{x\to 0}\frac{(-2x/e)}{\log(1+(-2x/e))}=\frac{1}{\lim_{y\to 0}\frac{\log(1+y)}{y}}=1$$

as was to be shown!

Mark Viola
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Hint. We have $$ x \to 0 \Longleftrightarrow -\frac{2x}{e} \to 0. $$

Olivier Oloa
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  • If $x \to 0 \Longleftrightarrow -\frac{2x}{e} \to 0$, then $\lim_{x\to 0} \frac{-\frac{2x}{e}}{\ln(1+(-\frac{2x}{e}))}=\frac{0}{\ln(1+0)}$? – Dave Dec 26 '16 at 16:56
  • If $\lim_{a\to 0} \frac{\ln(1+a)}{a}=1$, then in particular with $a:=-\frac{2x}e$ we get $\lim_{x\to 0} \frac{\ln(1+(-\frac{2x}{e}))}{-\frac{2x}{e}}=1.$ Are you Ok with this? – Olivier Oloa Dec 26 '16 at 16:57
  • Yes. But why $\lim_{x\to 0} \frac{-\frac{2x}{e}}{\ln(1+(-\frac{2x}{e}))}=1$? – Dave Dec 26 '16 at 17:06
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Hint: The function $f(x)=1/x$ is continuous, so $\lim_{n\to\infty}f(x_n)=f(\lim_{n\to\infty} x_n)$ as long as the sequence does not tend to zero.

Clayton
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