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

Question

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?

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

It's right?

Answer 1

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!

Answer 2

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

Answer 3

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.